id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-014501 | Real Analysis: Series — Parameter Sensitivity | 8 | Make each step logically reversible (or explain if not): Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{29}{16}}}.$$
(a) Solve using a named convergenc... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{29}{16}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-series te... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{29}{16}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014502 | Real Analysis: Series — p-Series Threshold | 8 | Find the exact value: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{52}{17}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cro... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{52}{17}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{52}{17}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity an... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{52}{17}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014503 | Real Analysis: Series — Integral Test for Power Laws | 8 | Warm-up: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{27}{7}}}.$$
(a) Solve using a named conver... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{27}{7}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{27}{7}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-series tes... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{27}{7}$, so the series is convergent. |
math-014504 | Real Analysis: Series — Integral Test for Power Laws | 8 | Give reasoning, not just computation: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{23}{4}}}.$$
(a) Solve... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{23}{4}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{23}{4}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were pert... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{23}{4}$, so the series is convergent. |
math-014505 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Solve (and briefly cross-validate): For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{9}{2}}}.$$
(a) Solve us... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{9}{2}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-se... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{9}{2}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014506 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Do not skip justification steps: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{116}{33}}}.$$
(a) Solve using a named convergence t... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{116}{33}$.",
"Final step: By the p-series test, it con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{116}{33}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{116}{33}$, so the series is convergent. |
math-014507 | Real Analysis: Series — p-Series Threshold | 8 | Solve and include a self-check: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{70}{19}}}.$$
(a) Solve using a named convergence test.
(b) Give an indep... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{70}{19}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{70}{19}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014508 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Exercise: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{68}{31}}}.$$
(a) Solve using a named convergence ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{68}{31}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-series te... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{68}{31}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014509 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Carefully track domains: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{74}{13}}}.$$
(a) Solve usi... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{74}{13}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-ser... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{74}{13}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014510 | Real Analysis: Series — p-Series Threshold | 8 | Complete the analysis: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{59}{6}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{59}{6}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: Th... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{59}{6}$, so the series is convergent. |
math-014511 | Real Analysis: Series — p-Series Threshold | 8 | Answer using clear logical steps: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{85}{37}}}.$$
(a) Solve us... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{85}{37}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{85}{37}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{85}{37}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014512 | Real Analysis: Series — Parameter Sensitivity | 8 | Answer with a short justification: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{75}{17}}}.$$
(a)... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{75}{17}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{75}{17}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{75}{17}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014513 | Real Analysis: Series — p-Series Threshold | 8 | Answer using clear logical steps: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{33}{10}}}.$$
(a) Solve using a named convergence test.
(b) Give an indep... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{33}{10}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{33}{10}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{33}{10}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014514 | Real Analysis: Series — Parameter Sensitivity | 8 | Solve with verification: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{29}{20}}}.$$
(a) Solve usi... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{29}{20}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{29}{20}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{29}{20}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014515 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Keep the final answer in boxed form: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{95}{21}}}.$$
(a) Solve using a named convergence test.
(b) Give an ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{95}{21}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{95}{21}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{95}{21}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014516 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Use two approaches if possible: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{43}{24}}}.$$
(a) Solve using a named convergence test.
(b) Give an indep... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{43}{24}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{43}{24}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{43}{24}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014517 | Real Analysis: Series — p-Series Threshold | 8 | Work this out carefully: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{31}{35}}}.$$
(a) Solve usi... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{31}{35}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{31}{35}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014518 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Provide both a computational and a conceptual explanation: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{115}{33}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a speci... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{115}{33}$, so the series is convergent. |
math-014519 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Solve with verification: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{72}{25}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{72}{25}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{72}{25}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014520 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Complete the analysis: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{71}{37}}}.$$
(a) Solve using... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{71}{37}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{71}{37}$, so the series is convergent. |
math-014521 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Where appropriate, name the theorem you use: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{39}{20}}}.$$
(a) Solve using a named co... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{39}{20}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{39}{20}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014522 | Real Analysis: Series — Integral Test for Power Laws | 8 | Solve and sanity-check: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{49}{10}}}.$$
(a) Solve using a name... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{49}{10}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{49}{10}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{49}{10}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014523 | Real Analysis: Series — Integral Test for Power Laws | 8 | Indicate where a theorem is used: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{6}{11}}}.$$
(a) Solve using a named convergence te... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{6}{11}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Robustness note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{6}{11}$, so the series is divergent. |
math-014524 | Real Analysis: Series — Integral Test for Power Laws | 8 | Work this out carefully: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{14}{3}}}.$$
(a) Solve usin... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{14}{3}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a special... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{14}{3}$, so the series is convergent. |
math-014525 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Use two approaches if possible: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{34}{29}}}.$$
(a) Solve using a named convergence tes... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{34}{29}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{34}{29}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014526 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Solve (and briefly cross-validate): Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{47}{8}}}.$$
(a) Solve using a named convergence ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{47}{8}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{47}{8}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-series tes... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{47}{8}$, so the series is convergent. |
math-014527 | Real Analysis: Series — Parameter Sensitivity | 8 | Complete the analysis: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{84}{29}}}.$$
(a) Solve using a named convergence test.
(b) Gi... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{84}{29}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{84}{29}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{84}{29}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014528 | Real Analysis: Series — Integral Test for Power Laws | 8 | Keep the final answer in boxed form: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{1}{4}}}.$$
(a) Solve using a named convergence ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{1}{4}$.",
"Final step: By the p-series test, it diverg... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{1}{4}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Generality note: The p-series test is a special... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{1}{4}$, so the series is divergent. |
math-014529 | Real Analysis: Series — p-Series Threshold | 8 | Write the solution set clearly: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{81}{31}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepen... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{81}{31}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{81}{31}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{81}{31}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014530 | Real Analysis: Series — Parameter Sensitivity | 8 | Find the exact value: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{71}{34}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cro... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{71}{34}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{71}{34}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014531 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Solve and sanity-check: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{67}{14}}}.$$
(a) Solve usin... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{67}{14}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{67}{14}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{67}{14}$, so the series is convergent. |
math-014532 | Real Analysis: Series — p-Series Threshold | 8 | Solve and include a self-check: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{103}{32}}}.$$
(a) S... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{103}{32}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{103}{32}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014533 | Real Analysis: Series — Integral Test for Power Laws | 8 | Question: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{35}{24}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross-check usi... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{35}{24}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-ser... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{35}{24}$, so the series is convergent. |
math-014534 | Real Analysis: Series — Integral Test for Power Laws | 8 | Exercise: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{16}{5}}}.$$
(a) Solve using a named convergence test.
(b) Give an independ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{16}{5}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test i... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{16}{5}$, so the series is convergent. |
math-014535 | Real Analysis: Series — p-Series Threshold | 8 | Keep the final answer in boxed form: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{13}{7}}}.$$
(a) Solve using a named convergence test.
(b) Give an ind... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{13}{7}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{13}{7}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014536 | Real Analysis: Series — p-Series Threshold | 8 | Try to avoid pattern-matching; explain why: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{19}{22}}}.$$
(a) Solve using a named con... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{19}{22}$.",
"Final step: By the p-series test, it dive... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{19}{22}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Robustness no... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{19}{22}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014537 | Real Analysis: Series — Integral Test for Power Laws | 8 | Explain what is being counted/optimized: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{10}{21}}}.$$
(a) Solve using a named conver... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{10}{21}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "If the problem were pe... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{10}{21}$, so the series is divergent. |
math-014538 | Real Analysis: Series — Integral Test for Power Laws | 8 | Complete the analysis: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{65}{12}}}.$$
(a) Solve using a named... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{65}{12}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were per... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{65}{12}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014539 | Real Analysis: Series — Integral Test for Power Laws | 8 | Compute the requested quantity: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{67}{23}}}.$$
(a) So... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{67}{23}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{67}{23}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{67}{23}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014540 | Real Analysis: Series — Integral Test for Power Laws | 8 | Track units/moduli carefully: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{55}{29}}}.$$
(a) Solve using a named convergence test.... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{55}{29}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{55}{29}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{55}{29}$, so the series is convergent. |
math-014541 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Give a theorem-based solution: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{69}{8}}}.$$
(a) Solve using a named convergence test.... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{69}{8}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{69}{8}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{69}{8}$, so the series is convergent. |
math-014542 | Real Analysis: Series — Integral Test for Power Laws | 8 | Solve and include a self-check: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{22}{3}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepe... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{22}{3}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{22}{3}$, so the series is convergent. |
math-014543 | Real Analysis: Series — Parameter Sensitivity | 8 | State any required conditions first: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{63}{38}}}.$$
(a) Solve using a named convergence test.
(b) Give an ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{63}{38}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{63}{38}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{63}{38}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014544 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Complete the analysis: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{43}{40}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cros... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{43}{40}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were per... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{43}{40}$, so the series is convergent. |
math-014545 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Work carefully and justify each inference: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{38}{29}}}.$$
(a) Solve using a named convergence test.
(b) Gi... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{38}{29}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{38}{29}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{38}{29}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014546 | Real Analysis: Series — Integral Test for Power Laws | 8 | Explain what is being counted/optimized: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{27}{5}}}.$$
(a) Solve using a named converg... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{27}{5}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity ana... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{27}{5}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014547 | Real Analysis: Series — Integral Test for Power Laws | 8 | Solve and justify each step: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{19}{8}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{19}{8}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{19}{8}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a special... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{19}{8}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014548 | Real Analysis: Series — p-Series Threshold | 8 | Start by stating any domain restrictions: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{18}{31}}}.... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{18}{31}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Generality note: The p-series test is a speci... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{18}{31}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014549 | Real Analysis: Series — Integral Test for Power Laws | 8 | Explain why your operations are valid: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{54}{35}}}.$$
(a) Solve using a named convergence test.
(b) Give a... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{54}{35}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{54}{35}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014550 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Carefully track domains: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{17}{40}}}.$$
(a) Solve using a named convergence test.
(b) ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{17}{40}$.",
"Final step: By the p-series test, it dive... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{17}{40}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Generality note: The p... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{17}{40}$, so the series is divergent. |
math-014551 | Real Analysis: Series — p-Series Threshold | 8 | Derive the result step-by-step: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{73}{29}}}.$$
(a) Solve using a named convergence tes... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{73}{29}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{73}{29}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{73}{29}$, so the series is convergent. |
math-014552 | Real Analysis: Series — Integral Test for Power Laws | 8 | Answer using clear logical steps: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{83}{18}}}.$$
(a) Solve using a named convergence test.
(b) Give an ind... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{83}{18}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{83}{18}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014553 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Explain each transformation: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{53}{33}}}.$$
(a) Solve... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{53}{33}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: T... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{53}{33}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014554 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Question: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{119}{19}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepe... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{119}{19}$.",
"Final step: By the p-series test, it con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{119}{19}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{119}{19}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014555 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Solve and then verify: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{51}{38}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cr... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{51}{38}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{51}{38}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-ser... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{51}{38}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014556 | Real Analysis: Series — p-Series Threshold | 8 | Work this out carefully: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{18}{7}}}.$$
(a) Solve usin... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{18}{7}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: Th... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{18}{7}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014557 | Real Analysis: Series — Integral Test for Power Laws | 8 | Provide both a computational and a conceptual explanation: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{98}{15}}}.$$
(a) Solve using a named converge... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{98}{15}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{98}{15}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014558 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Complete the analysis: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{20}{37}}}.$$
(a) Solve using a named... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{20}{37}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{20}{37}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014559 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Solve (and briefly cross-validate): Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{19}{30}}}.$$
(a) Solve using a named convergence... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{19}{30}$.",
"Final step: By the p-series test, it dive... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{19}{30}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{19}{30}$, so the series is divergent. |
math-014560 | Real Analysis: Series — Parameter Sensitivity | 8 | Determine the requested value: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{51}{23}}}.$$
(a) Solve using a named convergence test.
(b) Give an independ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{51}{23}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{51}{23}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014561 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Answer using clear logical steps: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{43}{23}}}.$$
(a) Solve using a named convergence t... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{43}{23}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{43}{23}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014562 | Real Analysis: Series — p-Series Threshold | 8 | Answer using clear logical steps: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{111}{31}}}.$$
(a) Solve using a named convergence test.
(b) Give an in... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{111}{31}$.",
"Final step: By the p-series test, it con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{111}{31}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a speci... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{111}{31}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014563 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Exercise: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{62}{33}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross-check usi... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{62}{33}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{62}{33}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014564 | Real Analysis: Series — Parameter Sensitivity | 8 | Solve with verification: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{93}{10}}}.$$
(a) Solve using a nam... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{93}{10}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were per... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{93}{10}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014565 | Real Analysis: Series — Divergence at the Boundary Case | 8 | State any required conditions first: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{118}{25}}}.$$
(a) Solve using a named convergence test.
(b) Give an... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{118}{25}$.",
"Final step: By the p-series test, it con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{118}{25}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{118}{25}$, so the series is convergent. |
math-014566 | Real Analysis: Series — Integral Test for Power Laws | 8 | Exercise: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{81}{25}}}.$$
(a) Solve using a named conv... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{81}{25}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{81}{25}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014567 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Make each step logically reversible (or explain if not): Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{18}{5}}}.$$
(a) Solve using... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{18}{5}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{18}{5}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a special... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{18}{5}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014568 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Carefully track domains: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{72}{17}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{72}{17}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{72}{17}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-series te... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{72}{17}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014569 | Real Analysis: Series — p-Series Threshold | 8 | Problem: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{5}{11}}}.$$
(a) Solve using a named convergence te... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{5}{11}$.",
"Final step: By the p-series test, it diver... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{5}{11}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Generality not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{5}{11}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014570 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Answer using clear logical steps: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{63}{17}}}.$$
(a) Solve using a named convergence t... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{63}{17}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{63}{17}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{63}{17}$, so the series is convergent. |
math-014571 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Provide both a computational and a conceptual explanation: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{61}{28}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{61}{28}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{61}{28}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014572 | Real Analysis: Series — Parameter Sensitivity | 8 | Carefully track domains: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{42}{23}}}.$$
(a) Solve using a named convergence test.
(b) ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{42}{23}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{42}{23}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{42}{23}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014573 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Warm-up: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{36}{13}}}.$$
(a) Solve using a named conve... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{36}{13}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{36}{13}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014574 | Real Analysis: Series — Parameter Sensitivity | 8 | Start by stating any domain restrictions: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{65}{24}}}.$$
(a) Solve using a named conve... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{65}{24}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{65}{24}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{65}{24}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014575 | Real Analysis: Series — Integral Test for Power Laws | 8 | Solve and then verify: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{37}{11}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cr... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{37}{11}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{37}{11}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014576 | Real Analysis: Series — Parameter Sensitivity | 8 | Solve (and briefly cross-validate): Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{11}{4}}}.$$
(a) Solve using a named convergence ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{11}{4}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{11}{4}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{11}{4}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014577 | Real Analysis: Series — Integral Test for Power Laws | 8 | Solve and sanity-check: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{2}{15}}}.$$
(a) Solve using a named convergence test.
(b) Gi... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{2}{15}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{2}{15}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014578 | Real Analysis: Series — Integral Test for Power Laws | 8 | Show all reasoning: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{53}{17}}}.$$
(a) Solve using a named convergence test.
(b) Give ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{53}{17}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{53}{17}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were per... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{53}{17}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014579 | Real Analysis: Series — Parameter Sensitivity | 8 | Derive the result step-by-step: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{73}{16}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepen... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{73}{16}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{73}{16}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{73}{16}$, so the series is convergent. |
math-014580 | Real Analysis: Series — Parameter Sensitivity | 8 | Be explicit about assumptions: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{61}{29}}}.$$
(a) Solve using a named convergence test.
(b) Give an independ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{61}{29}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{61}{29}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{61}{29}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014581 | Real Analysis: Series — p-Series Threshold | 8 | Where appropriate, name the theorem you use: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{23}{21}}}.$$
(a) Solve using a named co... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{23}{21}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{23}{21}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{23}{21}$, so the series is convergent. |
math-014582 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Track quantifiers carefully: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{35}{6}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{35}{6}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{35}{6}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{35}{6}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014583 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Explain what is being counted/optimized: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{8}{19}}}.$$
(a) Solve using a named convergence test.
(b) Give an... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{8}{19}$.",
"Final step: By the p-series test, it diver... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{8}{19}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Robustness not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{8}{19}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014584 | Real Analysis: Series — p-Series Threshold | 8 | Challenge: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{45}{38}}}.$$
(a) Solve using a named convergence... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{45}{38}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{45}{38}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were per... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{45}{38}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014585 | Real Analysis: Series — Integral Test for Power Laws | 8 | Do not skip justification steps: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{25}{38}}}.$$
(a) Solve usi... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{25}{38}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Sensitivity a... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{25}{38}$, so the series is divergent. |
math-014586 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Solve and include a self-check: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{17}{25}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepen... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{17}{25}$.",
"Final step: By the p-series test, it dive... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{17}{25}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Robustness note: The p-series test is a... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{17}{25}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014587 | Real Analysis: Series — Integral Test for Power Laws | 8 | Question: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{55}{24}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross-check using... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{55}{24}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{55}{24}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{55}{24}$, so the series is convergent. |
math-014588 | Real Analysis: Series — Integral Test for Power Laws | 8 | Solve and include a self-check: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{47}{31}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepen... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{47}{31}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{47}{31}$, so the series is convergent. |
math-014589 | Real Analysis: Series — Parameter Sensitivity | 8 | Write the solution set clearly: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{59}{7}}}.$$
(a) Solve using a named convergence test... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{59}{7}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{59}{7}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{59}{7}$, so the series is convergent. |
math-014590 | Real Analysis: Series — Integral Test for Power Laws | 8 | Explain each transformation: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{97}{33}}}.$$
(a) Solve... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{97}{33}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{97}{33}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-series te... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{97}{33}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014591 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Write the solution set clearly: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{7}{8}}}.$$
(a) Solv... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{7}{8}$.",
"Final step: By the p-series test, it diverg... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{7}{8}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Robustness note: The p-s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{7}{8}$, so the series is divergent. |
math-014592 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Question: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{34}{11}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepen... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{34}{11}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{34}{11}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014593 | Real Analysis: Series — Parameter Sensitivity | 8 | State any required conditions first: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{21}{19}}}.$$
(a) Solve using a named convergence test.
(b) Give an in... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{21}{19}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity an... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{21}{19}$, so the series is convergent. |
math-014594 | Real Analysis: Series — p-Series Threshold | 8 | Determine the requested value: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{99}{13}}}.$$
(a) Solve using... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{99}{13}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-ser... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{99}{13}$, so the series is convergent. |
math-014595 | Real Analysis: Series — p-Series Threshold | 8 | Solve and include a self-check: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{62}{21}}}.$$
(a) Solve usin... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{62}{21}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{62}{21}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity an... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{62}{21}$, so the series is convergent. |
math-014596 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Show all reasoning: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{84}{11}}}.$$
(a) Solve using a named co... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{84}{11}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{84}{11}$, so the series is convergent. |
math-014597 | Real Analysis: Series — Parameter Sensitivity | 8 | Give a fully justified solution: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{71}{38}}}.$$
(a) S... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{71}{38}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{71}{38}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014598 | Real Analysis: Series — Parameter Sensitivity | 8 | Explain why your operations are valid: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{13}{25}}}.$$
(a) Sol... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{13}{25}$.",
"Final step: By the p-series test, it dive... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{13}{25}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Robustness note: The p-series test is a... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{13}{25}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014599 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Solve and justify each step: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{72}{29}}}.$$
(a) Solve using a named convergence test.
(b) Give an independen... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{72}{29}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{72}{29}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: T... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{72}{29}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014600 | Real Analysis: Series — Integral Test for Power Laws | 8 | Challenge: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{14}{27}}}.$$
(a) Solve using a named convergence... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{14}{27}$.",
"Final step: By the p-series test, it dive... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{14}{27}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "If the proble... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{14}{27}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
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