id string | topic string | difficulty int64 | problem_statement string | solution_paths list | reconciliation dict | error_catalogue list | conceptual_takeaway string |
|---|---|---|---|---|---|---|---|
math-014401 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Make each step logically reversible (or explain if not): 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{49}{5}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{49}{5}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{49}{5}$, so the series is convergent. |
math-014402 | Real Analysis: Series — Integral Test for Power Laws | 8 | Keep the final answer in boxed form: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{10}{37}}}.$$
(a) Solve using a named convergence test.
(b) Give an ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{10}{37}$.",
"Final step: By the p-series test, it dive... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{10}{37}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "If the problem were perturbed: The p-se... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{10}{37}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014403 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Solve and sanity-check: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{19}{7}}}.$$
(a) Solve using a named... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{19}{7}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{19}{7}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{19}{7}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014404 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Give an answer and a quick verification: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{21}{29}}}.$... | [
{
"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 $p=\\frac{21}{29}$.",
"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 $p=\\frac{21}{29}$ 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 $p=\frac{21}{29}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014405 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Provide both a computational and a conceptual explanation: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{103}{13}}}.$$
(a) Solve u... | [
{
"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 $p=\\frac{103}{13}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: 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 $p=\frac{103}{13}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014406 | Real Analysis: Series — p-Series Threshold | 8 | Keep the final answer in boxed form: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{39}{25}}}.$$
(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": "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 $p=\\frac{39}{25}$ 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... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{39}{25}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014407 | Real Analysis: Series — Parameter Sensitivity | 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{37}{16}}}.$$
(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 $p=\\frac{37}{16}$.",
"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 $p=\\frac{37}{16}$ 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 $p=\frac{37}{16}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014408 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Explain each transformation: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{7}{10}}}.$$
(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 $p=\\frac{7}{10}$.",
"Final step: By the p-series test, it diver... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{7}{10}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"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 $p=\frac{7}{10}$, so the series is divergent. |
math-014409 | Real Analysis: Series — Parameter Sensitivity | 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{73}{15}}}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{73}{15}$ 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... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{73}{15}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014410 | Real Analysis: Series — Integral Test for Power Laws | 8 | Work carefully and justify each inference: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{37}{23}}}.$$
(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 $p=\\frac{37}{23}$.",
"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 $p=\\frac{37}{23}$ 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 $p=\frac{37}{23}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014411 | Real Analysis: Series — p-Series Threshold | 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{73}{39}}}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{73}{39}$ 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 $p=\frac{73}{39}$, so the series is convergent. |
math-014412 | Real Analysis: Series — Divergence at the Boundary Case | 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{9}{35}}}.$$
(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 $p=\\frac{9}{35}$.",
"Final step: By the p-series test, it diver... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{9}{35}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"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... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{9}{35}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014413 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Do not skip justification steps: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{48}{5}}}.$$
(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 $p=\\frac{48}{5}$.",
"Final step: By the p-series test, it conve... | {
"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 $p=\\frac{48}{5}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-seri... | [
{
"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 $p=\frac{48}{5}$, so the series is convergent. |
math-014414 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Compute the requested quantity: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{37}{7}}}.$$
(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 $p=\\frac{37}{7}$.",
"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 $p=\\frac{37}{7}$ 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 $p=\frac{37}{7}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014415 | Real Analysis: Series — Divergence at the Boundary Case | 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{43}{30}}}.$$
(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 $p=\\frac{43}{30}$ 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... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{43}{30}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014416 | Real Analysis: Series — Integral Test for Power Laws | 8 | Solve with verification: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{41}{38}}}.$$
(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 $p=\\frac{41}{38}$.",
"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 $p=\\frac{41}{38}$ 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... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{41}{38}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014417 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Indicate where a theorem is used: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{17}{8}}}.$$
(a) Solve using a named convergence test.
(b) Give an inde... | [
{
"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 $p=\\frac{17}{8}$.",
"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 $p=\\frac{17}{8}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"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 $p=\frac{17}{8}$, so the series is convergent. |
math-014418 | Real Analysis: Series — p-Series Threshold | 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{1}{20}}}.$$
(a) Solve using a named convergence test.
(b) G... | [
{
"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 $p=\\frac{1}{20}$ 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 $p=\frac{1}{20}$, so the series is divergent. |
math-014419 | Real Analysis: Series — Parameter Sensitivity | 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{7}{4}}}.$$
(a) Solve using a named convergence test.
(b) Give an independe... | [
{
"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 $p=\\frac{7}{4}$.",
"Final step: By the p-series test, it conver... | {
"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 $p=\\frac{7}{4}$ 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 $p=\frac{7}{4}$, so the series is convergent. |
math-014420 | Real Analysis: Series — p-Series Threshold | 8 | Give a theorem-based solution: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{29}{10}}}.$$
(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": "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 $p=\\frac{29}{10}$ 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 $p=\frac{29}{10}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014421 | Real Analysis: Series — Integral Test for Power Laws | 8 | Track quantifiers 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{67}{29}}}.$$
(a) Solve using a named convergence test.
... | [
{
"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 $p=\\frac{67}{29}$ 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... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{67}{29}$, so the series is convergent. |
math-014422 | Real Analysis: Series — Parameter Sensitivity | 8 | State any required conditions first: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{5}{23}}}.$$
(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": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{5}{23}$ 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... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{5}{23}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014423 | Real Analysis: Series — p-Series Threshold | 8 | Work carefully and justify each inference: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{71}{12}}}.$$
(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": "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 $p=\\frac{71}{12}$ 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... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{71}{12}$, so the series is convergent. |
math-014424 | Real Analysis: Series — Divergence at the Boundary Case | 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{98}{37}}}.$$
(a) So... | [
{
"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 $p=\\frac{98}{37}$ 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 $p=\frac{98}{37}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014425 | Real Analysis: Series — Integral Test for Power Laws | 8 | Explain what is being counted/optimized: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{54}{13}}}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{54}{13}$ 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 $p=\frac{54}{13}$, so the series is convergent. |
math-014426 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Try to avoid pattern-matching; explain why: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{16}{7}}}.$$
(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 $p=\\frac{16}{7}$.",
"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 $p=\\frac{16}{7}$ 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 $p=\frac{16}{7}$, so the series is convergent. |
math-014427 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Explain each transformation: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{7}{34}}}.$$
(a) Solve using a named convergence test.
(b) Give an independe... | [
{
"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 $p=\\frac{7}{34}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"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... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{7}{34}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014428 | Real Analysis: Series — Parameter Sensitivity | 8 | Explain each transformation: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{2}{9}}}.$$
(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 $p=\\frac{2}{9}$.",
"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 $p=\\frac{2}{9}$ 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... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{2}{9}$, so the series is divergent. |
math-014429 | Real Analysis: Series — Divergence at the Boundary Case | 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{89}{26}}}.$$
(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 $p=\\frac{89}{26}$.",
"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 $p=\\frac{89}{26}$ 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 $p=\frac{89}{26}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014430 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Explain each transformation: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{107}{11}}}.$$
(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": "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 $p=\\frac{107}{11}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"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... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{107}{11}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014431 | Real Analysis: Series — Parameter Sensitivity | 8 | Provide a rigorous solution: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{57}{22}}}.$$
(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 $p=\\frac{57}{22}$.",
"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 $p=\\frac{57}{22}$ 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... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{57}{22}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014432 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Track units/moduli carefully: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{40}{11}}}.$$
(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 $p=\\frac{40}{11}$.",
"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 $p=\\frac{40}{11}$ 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 $p=\frac{40}{11}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014433 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Derive the result step-by-step: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{49}{27}}}.$$
(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 $p=\\frac{49}{27}$.",
"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 $p=\\frac{49}{27}$ 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 $p=\frac{49}{27}$, so the series is convergent. |
math-014434 | Real Analysis: Series — p-Series Threshold | 8 | Use two approaches if possible: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{116}{27}}}.$$
(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 $p=\\frac{116}{27}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: 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... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{116}{27}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014435 | Real Analysis: Series — p-Series Threshold | 8 | Indicate where a theorem is used: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{31}{11}}}.$$
(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 $p=\\frac{31}{11}$.",
"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 $p=\\frac{31}{11}$ 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 $p=\frac{31}{11}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014436 | Real Analysis: Series — Parameter Sensitivity | 8 | Prompt: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{4}{9}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross-check using 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 $p=\\frac{4}{9}$.",
"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 $p=\\frac{4}{9}$ 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... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{4}{9}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014437 | Real Analysis: Series — p-Series Threshold | 8 | State any required conditions first: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{64}{13}}}.$$
(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 $p=\\frac{64}{13}$ 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 $p=\frac{64}{13}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014438 | Real Analysis: Series — Integral Test for Power Laws | 8 | Provide a rigorous solution: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{32}{5}}}.$$
(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 $p=\\frac{32}{5}$.",
"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 $p=\\frac{32}{5}$ 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... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{32}{5}$, so the series is convergent. |
math-014439 | Real Analysis: Series — Parameter Sensitivity | 8 | Checkpoint: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{4}{37}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross-check 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: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{4}{37}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"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 $p=\frac{4}{37}$, so the series is divergent. |
math-014440 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Explain why your operations are valid: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{54}{17}}}.$$
(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 $p=\\frac{54}{17}$.",
"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 $p=\\frac{54}{17}$ 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 $p=\frac{54}{17}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014441 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Task: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{82}{9}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross-check using a di... | [
{
"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 $p=\\frac{82}{9}$.",
"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 $p=\\frac{82}{9}$ 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... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{82}{9}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014442 | Real Analysis: Series — Integral Test for Power Laws | 8 | Challenge: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{11}{38}}}.$$
(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 $p=\\frac{11}{38}$.",
"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 $p=\\frac{11}{38}$ 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 $p=\frac{11}{38}$, so the series is divergent. |
math-014443 | Real Analysis: Series — p-Series Threshold | 8 | Compute the requested quantity: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{93}{17}}}.$$
(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 $p=\\frac{93}{17}$.",
"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 $p=\\frac{93}{17}$ 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 $p=\frac{93}{17}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014444 | Real Analysis: Series — Integral Test for Power Laws | 8 | Find the exact 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{95}{13}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross... | [
{
"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 $p=\\frac{95}{13}$.",
"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 $p=\\frac{95}{13}$ 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 $p=\frac{95}{13}$, so the series is convergent. |
math-014445 | 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{1}{17}}}.$$
(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": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{1}{17}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"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 $p=\frac{1}{17}$, so the series is divergent. |
math-014446 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Make each step logically reversible (or explain if not): 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{4}{3}$.",
"Final step: By the p-series test, it conver... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{4}{3}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-se... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{4}{3}$, so the series is convergent. |
math-014447 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Where appropriate, name the theorem you use: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{76}{35}}}.$$
(a) Solve using a named convergence test.
(b) Gi... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{76}{35}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{76}{35}$, so the series is convergent. |
math-014448 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Prompt: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{91}{19}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross-check using... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{91}{19}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-series te... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{91}{19}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014449 | Real Analysis: Series — Integral Test for Power Laws | 8 | Try to avoid pattern-matching; explain why: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{19}{26}}}.$$
(a) Solve using a named convergence test.
(b) Giv... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{19}{26}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Sensitivity analysis: ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{19}{26}$, so the series is divergent. |
math-014450 | Real Analysis: Series — p-Series Threshold | 8 | Prompt: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{47}{10}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross-check using a... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{47}{10}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{47}{10}$, so the series is convergent. |
math-014451 | Real Analysis: Series — p-Series Threshold | 8 | Solve and include a self-check: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{107}{32}}}.$$
(a) Solve using a named convergence test.
(b) Give an inde... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{107}{32}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{107}{32}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014452 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Provide a rigorous solution: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{17}{37}}}.$$
(a) Solve using a... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{17}{37}$.",
"Final step: By the p-series test, it dive... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{17}{37}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Generality note: The p-series test is a speci... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{17}{37}$, so the series is divergent. |
math-014453 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Provide both a computational and a conceptual explanation: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^... | [
{
"method_name": "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 $p=\\frac{109}{27}$.",
"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 $p=\\frac{109}{27}$ 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... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{109}{27}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014454 | Real Analysis: Series — Divergence at the Boundary Case | 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{19}{16}}}.$$
(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 $p=\\frac{19}{16}$.",
"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 $p=\\frac{19}{16}$ 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... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{19}{16}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014455 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Work carefully and justify each inference: 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.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{41}{9}}}.... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{41}{9}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a special... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{41}{9}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014456 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Derive the result step-by-step: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{61}{25}}}.$$
(a) So... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{61}{25}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{61}{25}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{61}{25}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014457 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Indicate where a theorem is used: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{115}{12}}}.$$
(a) Solve u... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{115}{12}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{115}{12}$, so the series is convergent. |
math-014458 | Real Analysis: Series — Integral Test for Power Laws | 8 | Make each step logically reversible (or explain if not): Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{41}{24}}}.$$
(a) Solve using a named convergence ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{41}{24}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{41}{24}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{41}{24}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014459 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Work this out carefully: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{87}{19}}}.$$
(a) Solve using a nam... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{87}{19}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{87}{19}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{87}{19}$, so the series is convergent. |
math-014460 | Real Analysis: Series — Integral Test for Power Laws | 8 | Solve and justify each step: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{105}{32}}}.$$
(a) Solve using ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{105}{32}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality no... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{105}{32}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014461 | Real Analysis: Series — Integral Test for Power Laws | 8 | Explain what is being counted/optimized: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{12}{35}}}.$$
(a) S... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{12}{35}$.",
"Final step: By the p-series test, it dive... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{12}{35}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Sensitivity analysis: ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{12}{35}$, so the series is divergent. |
math-014462 | Real Analysis: Series — Integral Test for Power Laws | 8 | Compute the requested quantity: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{1}{7}}}.$$
(a) Solv... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{1}{7}$.",
"Final step: By the p-series test, it diverg... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{1}{7}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Generality note: The p-series test is a s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{1}{7}$, so the series is divergent. |
math-014463 | Real Analysis: Series — p-Series Threshold | 8 | Solve (and briefly cross-validate): Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{9}{17}}}.$$
(a)... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{9}{17}$.",
"Final step: By the p-series test, it diver... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{9}{17}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Generality note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{9}{17}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014464 | Real Analysis: Series — Parameter Sensitivity | 8 | Write the solution set clearly: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{31}{30}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepen... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{31}{30}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{31}{30}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{31}{30}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014465 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Solve and then verify: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{38}{5}}}.$$
(a) Solve using ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{38}{5}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity ana... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{38}{5}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014466 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Explain why your operations are valid: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{33}{13}}}.$$
(a) Solve using a named convergence test.
(b) Give a... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{33}{13}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity an... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{33}{13}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014467 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Derive the result step-by-step: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{74}{27}}}.$$
(a) Solve using a named convergence tes... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{74}{27}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{74}{27}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: T... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{74}{27}$, so the series is convergent. |
math-014468 | Real Analysis: Series — p-Series Threshold | 8 | Start by stating any domain restrictions: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{80}{17}}}.$$
(a) Solve using a named convergence test.
(b) Give ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{80}{17}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{80}{17}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{80}{17}$, so the series is convergent. |
math-014469 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Track quantifiers carefully: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{69}{17}}}.$$
(a) Solve using a named convergence test.
(b) Give an independen... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{69}{17}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{69}{17}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014470 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Show all reasoning: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{33}{16}}}.$$
(a) Solve using a named convergence test.
(b) Give ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{33}{16}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-series te... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{33}{16}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014471 | Real Analysis: Series — p-Series Threshold | 8 | Proceed methodically: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{91}{10}}}.$$
(a) Solve using a named ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{91}{10}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{91}{10}$, so the series is convergent. |
math-014472 | Real Analysis: Series — p-Series Threshold | 8 | Give reasoning, not just computation: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{86}{15}}}.$$
(a) Solve using a named convergen... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{86}{15}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{86}{15}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{86}{15}$, so the series is convergent. |
math-014473 | Real Analysis: Series — p-Series Threshold | 8 | Try to avoid pattern-matching; explain why: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{4}{5}}}.$$
(a) ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{4}{5}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Robustness note: The p-s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{4}{5}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014474 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Show all reasoning: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{14}{9}}}.$$
(a) Solve using a named convergence test.
(b) Give a... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{14}{9}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{14}{9}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were pert... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{14}{9}$, so the series is convergent. |
math-014475 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Give a theorem-based solution: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{41}{39}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepe... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{41}{39}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{41}{39}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{41}{39}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014476 | Real Analysis: Series — Parameter Sensitivity | 8 | Solve and sanity-check: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{93}{34}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cro... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{93}{34}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{93}{34}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{93}{34}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014477 | Real Analysis: Series — p-Series Threshold | 8 | Determine the requested value: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{20}{9}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepen... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{20}{9}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: Th... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{20}{9}$, so the series is convergent. |
math-014478 | Real Analysis: Series — Integral Test for Power Laws | 8 | Provide both a computational and a conceptual explanation: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{37}{28}}}.$$
(a) Solve using a named converge... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{37}{28}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{37}{28}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity an... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{37}{28}$, so the series is convergent. |
math-014479 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Be explicit about assumptions: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{71}{35}}}.$$
(a) Solve using a named convergence test... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{71}{35}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{71}{35}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-ser... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{71}{35}$, so the series is convergent. |
math-014480 | Real Analysis: Series — Integral Test for Power Laws | 8 | Answer using clear logical steps: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{64}{9}}}.$$
(a) Solve using a named convergence test.
(b) Give an inde... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{64}{9}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{64}{9}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: Th... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{64}{9}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014481 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Give a fully justified solution: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{52}{21}}}.$$
(a) S... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{52}{21}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{52}{21}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{52}{21}$, so the series is convergent. |
math-014482 | Real Analysis: Series — p-Series Threshold | 8 | Track units/moduli carefully: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{2}{3}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{2}{3}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "Sensitivity analysis: Th... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{2}{3}$, so the series is divergent. |
math-014483 | Real Analysis: Series — Parameter Sensitivity | 8 | Do not skip justification steps: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{73}{24}}}.$$
(a) Solve usi... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{73}{24}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{73}{24}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014484 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Work carefully and justify each inference: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{31}{4}}}.$$
(a) Solve using a named convergence test.
(b) Give ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{31}{4}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{31}{4}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were perturbed: The p-seri... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{31}{4}$, so the series is convergent. |
math-014485 | Real Analysis: Series — Integral Test for Power Laws | 8 | Start by stating any domain restrictions: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{17}{18}}}.$$
(a) Solve using a named conve... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Diverges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{17}{18}$ is \\le 1, so both methods agree on \\boxed{\\text{Diverges}}.",
"robustness_analysis": "If the problem were perturbed: The p-se... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{17}{18}$, so the series is divergent. (Here the result is $\boxed{\text{Diverges}$.) |
math-014486 | Real Analysis: Series — Parameter Sensitivity | 8 | Show all reasoning: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{26}{9}}}.$$
(a) Solve using a named con... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{26}{9}$.",
"Final step: By the p-series test, it conve... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{26}{9}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{26}{9}$, so the series is convergent. |
math-014487 | Real Analysis: Series — Parameter Sensitivity | 8 | Solve and include a self-check: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{97}{11}}}.$$
(a) Solve using a named convergence test.
(b) Give an indepen... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{97}{11}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{97}{11}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014488 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Use two approaches if possible: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{83}{34}}}.$$
(a) Solve using a named convergence test.
(b) Give an indep... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{83}{34}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{83}{34}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014489 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Show all reasoning: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{48}{25}}}.$$
(a) Solve using a ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{48}{25}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{48}{25}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{48}{25}$, so the series is convergent. |
math-014490 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Explain each transformation: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{12}{5}}}.$$
(a) Solve using a ... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{12}{5}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a sp... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{12}{5}$, so the series is convergent. |
math-014491 | Real Analysis: Series — Integral Test for Power Laws | 8 | Provide a rigorous solution: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{39}{4}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{39}{4}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test i... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{39}{4}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014492 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Solve with verification: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{20}{19}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cr... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{20}{19}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: The p-series test is a s... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{20}{19}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014493 | Real Analysis: Series — Divergence at the Boundary Case | 8 | Solve with verification: Classify the infinite series (convergent or divergent). Provide two independent arguments (e.g., p-series test and integral test):
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{90}{29}}}.$$
(a) Solve usi... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{90}{29}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{90}{29}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity analysis: T... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{90}{29}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014494 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Show all reasoning: Decide convergence/divergence of the series and justify with a named theorem:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{88}{27}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{88}{27}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{88}{27}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality not... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Takeaway: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{88}{27}$, so the series is convergent. |
math-014495 | Real Analysis: Series — p-Series Threshold | 8 | Explain what is being counted/optimized: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{19}{4}}}.$$
(a) Solve using a named converg... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{19}{4}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Sensitivity ana... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{19}{4}$, so the series is convergent. |
math-014496 | Real Analysis: Series — Parameter Sensitivity | 8 | Find the exact value: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{43}{11}}}.$$
(a) Solve using a named convergence test.
(b) Giv... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "Both approaches agree after simplification. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{43}{11}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness note: The p-series test is a specia... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{43}{11}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014497 | Real Analysis: Series — Integral Test for Power Laws | 8 | Solve and then verify: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{85}{14}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cros... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{85}{14}$.",
"Final step: By the p-series test, it conv... | {
"consistency_check": "Cross-check: both derivations land on the same invariant quantity. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{85}{14}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "If the problem were per... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Remember: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{85}{14}$, so the series is convergent. |
math-014498 | Real Analysis: Series — Necessary vs Sufficient Conditions | 8 | Compute the requested quantity: For the parameterized p-series below, determine whether it converges. Your solution must reference the threshold at $p=1$:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{22}{19}}}.$$
(a) Solve usin... | [
{
"method_name": "Integral Test",
"approach": "Apply the integral test to $f(x)=x^{-p}$ and analyze $\\int_1^\\infty x^{-p}\\,dx$.",
"steps": [
"Step 1: Let $f(x)=x^{-p}$ for $x\\ge 1$. For $p>0$, $f$ is positive and decreasing.",
"Step 2: By the integral test, $\\sum_{n=1}^\\infty f(n)$ con... | {
"consistency_check": "The two methods are consistent and must coincide. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{22}{19}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Generality note: The p-series test is a ... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Core principle: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{22}{19}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014499 | Real Analysis: Series — Parameter Sensitivity | 8 | Task: Analyze the series using two distinct convergence tests and reconcile them:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{107}{24}}}.$$
(a) Solve using a named convergence test.
(b) Give an independent cross-check using a ... | [
{
"method_name": "p-Series Test",
"approach": "Use the theorem that $\\sum 1/n^p$ converges iff $p>1$.",
"steps": [
"Step 1: Recognize the given series as a p-series $\\sum_{n=1}^\\infty 1/n^p$ with $p>0$.",
"Step 2: Here $p=\\frac{107}{24}$.",
"Final step: By the p-series test, it con... | {
"consistency_check": "Consistency verification shows both paths yield the identical boxed result. Final answer: $\\boxed{\\text{Converges}$.\nBoth tests reduce to the same threshold $p=1$. Here $p=\\frac{107}{24}$ is > 1, so both methods agree on \\boxed{\\text{Converges}}.",
"robustness_analysis": "Robustness no... | [
{
"error_description": "Concluded convergence because $1/n^p\\to 0$.",
"why_plausible": "The term test (terms go to 0) is often overused as if it were sufficient.",
"why_wrong": "While $a_n\\to 0$ is necessary, it is not sufficient (e.g., $\\sum 1/n$ diverges).",
"which_method_catches_it": "Integral... | Key idea: For the p-series $\sum 1/n^p$, convergence is controlled entirely by whether $p>1$. Here $p=\frac{107}{24}$, so the series is convergent. (Here the result is $\boxed{\text{Converges}$.) |
math-014500 | Real Analysis: Series — Parameter Sensitivity | 8 | Find the exact value: Is the following series convergent? Give one theorem-level reason and one alternative validation:
Decide whether the series converges or diverges. You must justify your answer by naming a theorem.
$$\sum_{n=1}^\infty \frac{1}{n^{\frac{23}{6}}}.$$
(a) Solve using a named convergence test.
(b) Give... | [
{
"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 $p=\\frac{23}{6}$ 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 $p=\frac{23}{6}$, so the series is convergent. |
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