Comparison Tests for Convergence - AP Calculus BC

An alternating-free series is $\sum_{n=1}^{\infty} \frac{1}{n^2 + \ln n}$ . Which comparison correctly determines convergence?
Answer: Since $\frac{1}{n^2+\ln n}\le \frac{1}{n^2}$ for $n\ge1$ and $\sum \frac{1}{n^2}$ converges, the series converges. This problem tests comparison with a logarithmic perturbation. Since $\ln n > 0$ for $n \geq 2$ , we have $n^2 + \ln n > n^2$ , which means $\frac{1}{n^2 + \ln n} < \frac{1}{n^2}$ . The series $\sum \frac{1}{n^2}$ converges (p-series with $p = 2 > 1$ ), so by comparison, our series converges. Choice A incorrectly reverses the inequality—adding positive terms to the denominator decreases the fraction value. When logarithmic terms appear with polynomial terms, the polynomial behavior dominates for large n, so you can often ignore the logarithmic term when choosing your comparison series.
Use comparison to determine whether $\sum_{n=1}^{\infty} \frac{n}{n^2+\sin n}$ converges.
Answer: Diverges by limit comparison with $\sum \frac{1}{n}$ since $\lim_{n\to\infty}\frac{\frac{n}{n^2+\sin n}}{\frac{1}{n}}=1$ . The skill here is using comparison tests to determine the convergence of infinite series. To analyze the series ∑ n/( $n^2$ + sin n), note that sin n is bounded between -1 and 1, so the denominator is between $n^2$ - 1 and $n^2$ + 1, making the term like n/ $n^2$ = 1/n. Apply the limit comparison test with $b_n$ = 1/n, computing lim [n/( $n^2$ + sin n)] / (1/n) = lim $n^2$ /( $n^2$ + sin n) = 1, positive and finite. Since ∑ 1/n diverges, the given series diverges. A tempting distractor is suggesting convergence by limit comparison with ∑ 1/n where the limit is 0, but the actual limit is 1, not 0. A transferable comparison strategy is to bound oscillating terms like sin n and use limit comparison with the dominant polynomial behavior.
To estimate error, a sensor uses $sum_{n=1}^{infty}frac{3n+1}{n^3+2}$ . Which comparison correctly determines convergence?
Answer: Since $\frac{3n+1}{n^3+2}\le \frac{4}{n^2}$ for $n\ge1$ and $\sum \frac{4}{n^2}$ converges, the series converges. This problem requires using the comparison test to determine convergence of the series. For large n, the dominant terms are 3n in the numerator and n³ in the denominator, so the series behaves like 3/n². Since n³ + 2 > n³, we have (3n+1)/(n³+2) < (3n+1)/n³ < 4n/n³ = 4/n² for n ≥ 1. The series ∑4/n² = 4∑1/n² converges (p-series with p = 2 > 1), so by the comparison test, our series converges. Choice A incorrectly reverses the inequality direction, which would require the divergence of the comparison series to conclude anything. When using the comparison test, always verify your inequality direction matches your conclusion: if your series is less than a convergent series, it converges.
A power-loss estimate uses $\sum_{n=1}^{\infty} \frac{n}{n^3+7n+1}$ . Which comparison is valid?
Answer: Since $\frac{n}{n^3+7n+1}\le \frac{n}{n^3}=\frac{1}{n^2}$ for $n\ge1$ and $\sum \frac{1}{n^2}$ converges, the series converges. This problem tests comparison for a rational function series. The dominant behavior is $n/n^3 = 1/n^2$ for large n. Since $n^3 + 7n + 1 > n^3$ , we have $n/(n^3+7n+1) < n/n^3 = 1/n^2$ . The series $\sum 1/n^2$ converges (p-series with p = 2 > 1), so our series converges by comparison. Choice A incorrectly states the inequality in the wrong direction, and even if it were correct, we couldn't conclude convergence from our series being larger than a convergent series. For rational functions, identify the highest degree terms in numerator and denominator to determine the comparison series, then verify the inequality direction.
Determine whether $\sum_{n=1}^{\infty} \frac{2^n}{3^n+n}$ converges by comparison.
Answer: Converges by comparison with $\sum \left(\frac{2}{3}\right)^n$ since $\frac{2^n}{3^n+n}<\left(\frac{2}{3}\right)^n$ for all $n$ . The skill here is using comparison tests to determine the convergence of infinite series. To analyze the series ∑ $2^n$ /( $3^n$ + n), observe that for large n, the term is less than $2^n$ / $3^n$ = $(2/3)^n$ since the denominator is larger than $3^n$ . Apply the direct comparison test with $b_n$ = $(2/3)^n$ , as $2^n$ /( $3^n$ + n) < $(2/3)^n$ for all n ≥ 1. Since ∑ $(2/3)^n$ is a geometric series with ratio <1 and converges, the given series converges. A tempting distractor is claiming divergence by limit comparison with ∑ $(2/3)^n$ where the limit is 0, but a limit of 0 with a convergent series does not imply divergence. A transferable comparison strategy is to use direct comparison with geometric series when exponential terms dominate, ensuring the inequality holds for all terms.
Which comparison correctly determines whether $\sum_{n=1}^\infty \frac{1}{n^2+\sin n}$ converges?
Answer: Since $\frac{1}{n^2+\sin n}\le \frac{1}{n^2}$ for $n\ge1$ and $\sum \frac{1}{n^2}$ converges, the series converges. This problem involves comparison tests with bounded oscillating terms. For $\sum_{n=1}^\infty \frac{1}{n^2+\sin n}$ , note that $-1 \leq \sin n \leq 1$ for all $n$ . This gives us $n^2-1 \leq n^2+\sin n \leq n^2+1$ , which means $\frac{1}{n^2+1} \leq \frac{1}{n^2+\sin n} \leq \frac{1}{n^2-1}$ for $n \geq 2$ . Since both $\sum \frac{1}{n^2+1}$ and $\sum \frac{1}{n^2-1}$ behave like $\sum \frac{1}{n^2}$ , which converges, our series converges by comparison. Choice B correctly identifies this with the simpler bound $\frac{1}{n^2+\sin n} \leq \frac{1}{n^2-1} < \frac{2}{n^2}$ for $n \geq 2$ . Choice C incorrectly reverses the inequality. When dealing with bounded oscillating terms added to powers of $n$ , they don't affect the convergence behavior for large $n$ .
A data correction term is $sum_{n=1}^{infty}frac{\sqrt{n}}{n^2+1}$ . Which comparison correctly decides convergence?
Answer: Since $\frac{\sqrt{n}}{n^2+1}\le \frac{\sqrt{n}}{n^2}=\frac{1}{n^{3/2}}$ for $n\ge1$ and $\sum \frac{1}{n^{3/2}}$ converges, the series converges. This problem involves comparing a series with mixed powers of n. The term √n/(n²+1) behaves like √n/n² = 1/ $n^{3/2}$ for large n. Since n² + 1 > n², we have √n/(n²+1) < √n/n² = 1/ $n^{3/2}$ . The series ∑1/ $n^{3/2}$ converges (p-series with p = 3/2 > 1), so by comparison, our series converges. Choice B incorrectly reverses the inequality and draws the wrong conclusion from a convergent comparison. When simplifying expressions with roots and powers, combine exponents carefully: √n = $n^{1/2}$ , so √n/n² = $n^{1/2-2}$ = $n^{-3/2}$ = 1/ $n^{3/2}$ .
Use comparison to determine whether $\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n^3+n}}$ converges.
Answer: Since $\dfrac{1}{\sqrt{n^3+n}}\le \dfrac{1}{n^{3/2}}$ for $n\ge1$ and $\sum \dfrac{1}{n^{3/2}}$ converges, the series converges. The skill being tested is the comparison test for determining series convergence. To apply it to sum 1/√( $n^3$ + n), note the term behaves like 1/ $n^{3/2}$ for large n. Choose to compare with sum 1/ $n^{3/2}$ , verifying that 1/√( $n^3$ + n) ≤ 1/√( $n^3$ ) = 1/ $n^{3/2}$ for n ≥ 1 since the denominator is larger. Since sum 1/ $n^{3/2}$ converges (p=3/2>1), the original series converges by comparison. A tempting distractor is choice C, which claims $a_n$ ≥ 1/ $n^{3/2}$ and concludes divergence, but the inequality direction is reversed, invalidating the argument. A transferable comparison strategy is to ignore smaller positive terms in the denominator to establish an upper bound for convergence.
Decide whether $\sum_{n=1}^{\infty} \dfrac{2+\sin n}{n^2}$ converges using a comparison test.
Answer: Since $\dfrac{2+\sin n}{n^2}\le \dfrac{3}{n^2}$ and $\sum \dfrac{3}{n^2}$ converges, the series converges. The skill being tested is the comparison test for determining series convergence. To apply it to sum (2 + sin n)/ $n^2$ , bound the oscillating term since |sin n| ≤ 1, so 2 + sin n ≤ 3. Choose to compare with sum 3/ $n^2$ , as (2 + sin n)/ $n^2$ ≤ 3/ $n^2$ for all n. Since sum 3/ $n^2$ converges as a p-series with p=2>1, the original series converges by comparison. A tempting distractor is choice C, which uses $a_n$ ≤ 1/ $n^2$ but incorrectly states sum 1/ $n^2$ diverges and concludes divergence, confusing convergence properties. A transferable comparison strategy is to use absolute bounds on bounded oscillating parts to compare with scaled known series.
Determine convergence of $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2+1}$ using comparison.
Answer: Converges by limit comparison with $\sum \frac{1}{n^{3/2}}$ since $\lim_{n\to\infty}\frac{\frac{\sqrt{n}}{n^2+1}}{\frac{1}{n^{3/2}}}=1$ . The skill here is using comparison tests to determine the convergence of infinite series. To analyze the series ∑ √n/( $n^2$ + 1), observe that for large n, it behaves like √n/ $n^2$ = 1/ $n^{3/2}$ . Apply the limit comparison test with $b_n$ = 1/ $n^{3/2}$ , computing lim [√n/( $n^2$ + 1)] / ( $1/n^{3/2}$ ) = lim $n^2$ /( $n^2$ + 1) = 1, a positive finite value. Since ∑ 1/ $n^{3/2}$ converges (p = 3/2 > 1), the given series converges. A tempting distractor is claiming divergence by limit comparison with ∑ 1/n where the limit is 0, but a limit of 0 with a divergent series does not prove divergence. A transferable comparison strategy is to match the degree of the polynomial terms in numerator and denominator to identify the equivalent p-series for limit comparison.
Determine whether $\sum_{n=1}^{\infty} \frac{3n+1}{n^3+5}$ converges using a comparison test.
Answer: Converges by limit comparison with $\sum \frac{1}{n^2}$ since $\lim_{n\to\infty}\frac{\frac{3n+1}{n^3+5}}{\frac{1}{n^2}}=3$ . The skill here is using comparison tests to determine the convergence of infinite series. To analyze the series ∑ (3n+1)/( $n^3+5$ ), observe that for large n, the dominant terms are 3n/ $n^3$ = 3/ $n^2$ , suggesting similarity to the p-series ∑ 1/ $n^2$ . Apply the limit comparison test by computing lim [(3n+1)/( $n^3+5$ )] / ( $1/n^2$ ) = lim (3n+1) $n^2$ / ( $n^3+5$ ) = 3, which is positive and finite. Since ∑ 1/ $n^2$ converges, the given series also converges. A tempting distractor is claiming divergence by direct comparison with ∑ 1/n, but actually (3n+1)/( $n^3+5$ ) < 1/n for large n, and comparing a smaller series to a divergent one does not prove divergence. A transferable comparison strategy is to identify the asymptotic behavior of the general term and use limit comparison with the corresponding known series.
A model sums $sum_{n=2}^{infty}frac{\ln n}{n^{3/2}}$ . Which comparison test statement is valid?
Answer: Since $\frac{\ln n}{n^{3/2}}\le \frac{n^{1/4}}{n^{3/2}}=\frac{1}{n^{5/4}}$ for $n\ge2$ and $\sum \frac{1}{n^{5/4}}$ converges, the series converges. This problem tests applying the comparison test to a series involving logarithms. Since ln n grows slower than any positive power of n, we need to find an appropriate bound for ln n/ $n^{3/2}$ . For large n, ln n < $n^{1/4}$ (this can be verified by noting that ln n/ $n^{1/4}$ → 0 as n → ∞). Therefore, ln n/ $n^{3/2}$ < $n^{1/4}$ / $n^{3/2}$ = 1/ $n^{5/4}$ for sufficiently large n. Since ∑1/ $n^{5/4}$ converges (p-series with p = 5/4 > 1), our series converges by the comparison test. Choice B incorrectly uses a divergent comparison series with the wrong inequality direction. Remember that logarithmic growth is slower than any positive power growth, making comparisons with n^α for small positive α useful.
A simulation accumulates $sum_{n=1}^{infty}\frac{5}{n^2+4n}$ . Which comparison correctly concludes convergence?
Answer: Since $\frac{5}{n^2+4n}\le \frac{5}{n^2}$ for $n\ge1$ and $\sum \frac{5}{n^2}$ converges, the series converges. This problem involves a straightforward rational function comparison. Since n² + 4n > n² for positive n, we have 5/(n²+4n) < 5/n². The series ∑5/n² = 5∑1/n² converges because ∑1/n² is a convergent p-series with p = 2 > 1. By the comparison test, our series converges. Choice A incorrectly reverses the inequality—adding positive terms to the denominator makes the fraction smaller, not larger. When constants appear in series, factor them out: the convergence of ∑c·aₙ depends only on whether ∑aₙ converges (for c ≠ 0).
In a smoothing algorithm, consider $sum_{n=2}^{infty}frac{1}{n(\ln n)^2}$ . Which comparison statement is correct?
Answer: Since $\frac{1}{n(\ln n)^2}\le \frac{1}{n^{3/2}}$ for $n\ge2$ and $\sum \frac{1}{n^{3/2}}$ converges, the series converges. This problem tests comparison with logarithmic series, which require special care. The series ∑1/(n ln n) diverges by the integral test, but 1/(n(ln n)²) < 1/(n ln n), so we cannot conclude convergence from this comparison. However, for large n, (ln n)² grows much slower than $n^{1/2}$ , so 1/(n(ln n)²) < 1/ $n^{3/2}$ for sufficiently large n. Since ∑1/ $n^{3/2}$ converges (p-series with p = 3/2 > 1), our series converges. Choice B incorrectly tries to use a divergent comparison series with the series being smaller. When dealing with logarithmic terms, remember that powers of ln n grow slower than any positive power of n, making polynomial comparisons effective.
Determine convergence of $\sum_{n=1}^{\infty} \frac{n+4}{\sqrt{n^5+2}}$ using comparison.
Answer: Converges by limit comparison with $\sum \frac{1}{n^{3/2}}$ since $\lim_{n\to\infty}\frac{\frac{n+4}{\sqrt{n^5+2}}}{\frac{1}{n^{3/2}}}=1$ . The skill here is using comparison tests to determine the convergence of infinite series. To analyze the series ∑ (n+4)/√( $n^5$ + 2), observe that for large n, √( $n^5$ + 2) ≈ $n^{5/2}$ , so the term is like n / $n^{5/2}$ = 1/ $n^{3/2}$ . Apply the limit comparison test with $b_n$ = 1/ $n^{3/2}$ , finding lim [(n+4)/√( $n^5$ + 2)] / ( $1/n^{3/2}$ ) = 1, positive and finite. Since ∑ 1/ $n^{3/2}$ converges, the given series converges. A tempting distractor is claiming divergence by limit comparison with ∑ 1/n where the limit is 1, but the actual limit with 1/n is 0, not 1, and even then it wouldn't prove divergence. A transferable comparison strategy is to approximate square roots of polynomials by the leading term and use limit comparison with the resulting power.
Use comparison to determine whether $\sum_{n=1}^{\infty} \frac{n^2}{2n^3+7}$ converges.
Answer: Diverges by limit comparison with $\sum \frac{1}{n}$ since $\lim_{n\to\infty}\frac{\frac{n^2}{2n^3+7}}{\frac{1}{n}}=\frac{1}{2}$ . The skill here is using comparison tests to determine the convergence of infinite series. To analyze the series ∑ $n^2$ /( $2n^3$ + 7), note that for large n, it simplifies to $n^2$ /( $2n^3$ ) = 1/(2n). Apply the limit comparison test with $b_n$ = 1/n, finding lim [ $n^2$ /( $2n^3$ + 7)] / (1/n) = lim $n^3$ /( $2n^3$ + 7) = 1/2, positive and finite. Since ∑ 1/n diverges, the given series diverges. A tempting distractor is suggesting convergence by limit comparison with ∑ 1/n where the limit is 0, but the actual limit is 1/2, not 0, making that incorrect. A transferable comparison strategy is to always compute the limit precisely and ensure the reference series matches the dominant term's behavior.
Which comparison test correctly determines whether $\sum_{n=1}^\infty \frac{n^2}{n^4+1}$ converges?
Answer: Since $\frac{n^2}{n^4+1}\le \frac{1}{n^2}$ for $n\ge1$ and $\sum \frac{1}{n^2}$ converges, the series converges. This problem tests comparison test skills for rational function series. For $\sum_{n=1}^\infty \frac{n^2}{n^4+1}$ , we analyze the dominant behavior: as $n \to \infty$ , the terms behave like $\frac{n^2}{n^4} = \frac{1}{n^2}$ . To verify $\frac{n^2}{n^4+1} \leq \frac{1}{n^2}$ , we need $n^2 \cdot n^2 \leq n^4+1$ , which gives $n^4 \leq n^4+1$ , true for all $n$ . Since $\sum \frac{1}{n^2}$ is a convergent p-series (p=2>1), our series converges by direct comparison. Choice A incorrectly claims the opposite inequality, which would require $n^4 \geq n^4+1$ , clearly false. When dealing with rational functions, the convergence behavior is determined by the difference in the degrees of denominator and numerator: convergence occurs when this difference exceeds 1.
Which comparison test correctly determines whether $\sum_{n=1}^\infty \frac{\ln(n+1)}{n^2}$ converges?
Answer: Since $\frac{\ln(n+1)}{n^2}\le \frac{1}{n^{3/2}}$ for $n\ge1$ and $\sum \frac{1}{n^{3/2}}$ converges, the series converges. This problem tests comparison techniques for series involving logarithms. For $\sum_{n=1}^\infty \frac{\ln(n+1)}{n^2}$ , we need to bound the logarithmic growth. Since $\ln(n+1)$ grows slower than any positive power of $n$ , we can show that $\ln(n+1) < n^{1/2}$ for sufficiently large $n$ . This gives us $\frac{\ln(n+1)}{n^2} < \frac{n^{1/2}}{n^2} = \frac{1}{n^{3/2}}$ for large $n$ . Since $\sum \frac{1}{n^{3/2}}$ is a convergent p-series (p=3/2>1), our series converges by direct comparison. Choice A incorrectly claims $\frac{\ln(n+1)}{n^2} \geq \frac{1}{n}$ , which would require $\ln(n+1) \geq n$ , failing for all $n$ . Remember that logarithmic functions grow slower than any positive power of $n$ , making them negligible in comparison tests for large $n$ .
Consider $sum_{n=1}^{infty}frac{n^2+5}{n^4-1}$ from a computation routine. Which comparison justifies convergence?
Answer: Since $\frac{n^2+5}{n^4-1}\le \frac{2n^2}{n^4/2}=\frac{4}{n^2}$ for $n\ge2$ and $\sum \frac{4}{n^2}$ converges, the series converges. This problem involves comparing a rational function series for convergence. The dominant behavior for large n is n²/n⁴ = 1/n². Since n⁴ - 1 < n⁴, we have (n²+5)/(n⁴-1) > (n²+5)/n⁴. Also, since n⁴ - 1 > n⁴/2 for n ≥ 2, we get (n²+5)/(n⁴-1) < (n²+5)/(n⁴/2) = 2(n²+5)/n⁴ < 2(2n²)/n⁴ = 4/n² for large n. Since ∑4/n² converges, our series converges by comparison. Choice A incorrectly concludes divergence from a convergent comparison series being smaller than our series. When dealing with rational functions, focus on the highest power terms to determine the dominant behavior and appropriate comparison series.
Which comparison test correctly determines whether $\sum_{n=1}^\infty \frac{\sqrt{n}}{n^2+1}$ converges?
Answer: Since $\frac{\sqrt{n}}{n^2+1}\le \frac{1}{n^{3/2}}$ for $n\ge1$ and $\sum \frac{1}{n^{3/2}}$ converges, the series converges. This problem tests comparison test skills for series convergence. The series $\sum_{n=1}^\infty \frac{\sqrt{n}}{n^2+1}$ has terms that behave like $\frac{\sqrt{n}}{n^2} = \frac{1}{n^{3/2}}$ for large $n$ . To verify the inequality $\frac{\sqrt{n}}{n^2+1} \leq \frac{1}{n^{3/2}}$ , we need $\sqrt{n} \cdot n^{3/2} \leq n^2+1$ , which simplifies to $n^2 \leq n^2+1$ , clearly true for all $n \geq 1$ . Since $\sum \frac{1}{n^{3/2}}$ is a convergent p-series (p=3/2>1), our series converges by direct comparison. Choice A incorrectly reverses the inequality direction, which would lead to an inconclusive comparison. When choosing comparison series, match the growth rate of your terms and use p-series $\sum \frac{1}{n^p}$ as benchmarks, remembering they converge when $p>1$ .

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