AP Calculus BC Quiz: Comparison Tests For Convergence

What this quiz covers

This quiz focuses on Comparison Tests For Convergence, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Since $\frac{3n+1}{n^3+2}\ge \frac{1}{n^2}$ for $n\ge1$ and $\sum \frac{1}{n^2}$ converges, the series converges.
2. 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. (correct answer)
3. Since $\frac{3n+1}{n^3+2}\le \frac{1}{n}$ for $n\ge1$ and $\sum \frac{1}{n}$ diverges, the series converges.
4. Since $\frac{3n+1}{n^3+2}\ge \frac{1}{n^3}$ for $n\ge1$ and $\sum \frac{1}{n^3}$ converges, the series diverges.
5. Since $\frac{3n+1}{n^3+2}\le \frac{1}{n^3}$ for $n\ge1$ and $\sum \frac{1}{n^3}$ converges, the series diverges.

Explanation: 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.

Question 2

1. Since $\frac{\ln n}{n^{3/2}}\ge \frac{1}{n^{3/2}}$ for $n\ge2$ and $\sum \frac{1}{n^{3/2}}$ converges, the series diverges.
2. Since $\frac{\ln n}{n^{3/2}}\le \frac{1}{n}$ for $n\ge2$ and $\sum \frac{1}{n}$ diverges, the series converges.
3. 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. (correct answer)
4. Since $\frac{\ln n}{n^{3/2}}\ge \frac{1}{n^{5/4}}$ for $n\ge2$ and $\sum \frac{1}{n^{5/4}}$ converges, the series diverges.
5. Since $\frac{\ln n}{n^{3/2}}\le \frac{1}{n^{3/2}}$ for $n\ge2$ and $\sum \frac{1}{n^{3/2}}$ converges, the series diverges.

Explanation: 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.

Question 3

1. Since $\frac{1}{n^2+\ln n}\ge \frac{1}{n^2}$ for $n\ge1$ and $\sum \frac{1}{n^2}$ converges, the series converges.
2. Since $\frac{1}{n^2+\ln n}\le \frac{1}{n}$ for $n\ge1$ and $\sum \frac{1}{n}$ diverges, the series converges.
3. 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. (correct answer)
4. Since $\frac{1}{n^2+\ln n}\ge \frac{1}{n}$ for $n\ge1$ and $\sum \frac{1}{n}$ diverges, the series converges.
5. Since $\frac{1}{n^2+\ln n}\le \frac{1}{n^3}$ for $n\ge1$ and $\sum \frac{1}{n^3}$ converges, the series diverges.

Explanation: 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.

Question 4

1. Diverges by comparison with $\sum \frac{1}{n}$ since $\frac{\ln(n+1)}{n^2}>\frac{1}{n}$ for large $n$.
2. Converges by comparison with $\sum \frac{1}{n}$ since $\frac{\ln(n+1)}{n^2}<\frac{1}{n}$ for all $n$.
3. Converges by comparison with $\sum \frac{1}{n^{3/2}}$ since $\frac{\ln(n+1)}{n^2}<\frac{1}{n^{3/2}}$ for large $n$. (correct answer)
4. Diverges by limit comparison with $\sum \frac{1}{n^2}$ since $\lim_{n\to\infty}\frac{\frac{\ln(n+1)}{n^2}}{\frac{1}{n^2}}=\infty$.
5. Converges by limit comparison with $\sum \frac{1}{n}$ since $\lim_{n\to\infty}\frac{\frac{\ln(n+1)}{n^2}}{\frac{1}{n}}=1$.

Explanation: The skill here is using comparison tests to determine the convergence of infinite series. To analyze the series $\sum \frac{\ln(n+1)}{n^2}$, note that $\ln(n+1)$ grows slower than any positive power of $n$, such as $n^{1/2}$. For large $n$, $\ln(n+1) < n^{1/2}$, so $\frac{\ln(n+1)}{n^2} < \frac{n^{1/2}}{n^2} = \frac{1}{n^{3/2}}$. Apply the direct comparison test with $b_n = \frac{1}{n^{3/2}}$, which converges ($p=\frac{3}{2}>1$), so the series converges. A tempting distractor is claiming divergence by limit comparison with $\sum \frac{1}{n^2}$ where the limit is $\infty$ (actually $\lim \frac{\ln(n+1)}{n^2} \cdot n^2 = \infty$, but with convergent reference, it does not imply divergence). A transferable comparison strategy is to bound logarithmic terms with polynomial terms to enable direct comparison with a known p-series.

Question 5

1. Diverges by comparison with $\sum \frac{1}{\sqrt{n}}$ since $\frac{\sqrt{n}}{n^2+1}>\frac{1}{\sqrt{n}}$ for large $n$.
2. Converges by limit comparison with $\sum \frac{1}{n^{3/2}}$ since $\lim_{n\to\infty}\frac{\frac{\sqrt{n}}{n^2+1}}{\frac{1}{n^{3/2}}}=1$. (correct answer)
3. Diverges by limit comparison with $\sum \frac{1}{n}$ since $\lim_{n\to\infty}\frac{\frac{\sqrt{n}}{n^2+1}}{\frac{1}{n}}=0$.
4. Converges by comparison with $\sum \frac{1}{n}$ since $\frac{\sqrt{n}}{n^2+1}<\frac{1}{n}$ for all $n$.
5. Diverges by comparison with $\sum \frac{1}{n^2}$ since $\frac{\sqrt{n}}{n^2+1}>\frac{1}{n^2}$ for large $n$.

Explanation: 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.

Question 6

1. Since $\dfrac{\sqrt{n}}{n^2+1}\ge \dfrac{1}{n}$ and $\sum \dfrac{1}{n}$ diverges, the series converges.
2. Since $\dfrac{\sqrt{n}}{n^2+1}\le \dfrac{1}{n^{3/2}}$ for $n\ge1$ and $\sum \dfrac{1}{n^{3/2}}$ converges, the series converges. (correct answer)
3. Since $\dfrac{\sqrt{n}}{n^2+1}\ge \dfrac{1}{n^{3/2}}$ and $\sum \dfrac{1}{n^{3/2}}$ converges, the series diverges.
4. Since $\dfrac{\sqrt{n}}{n^2+1}\le \dfrac{1}{n}$ and $\sum \dfrac{1}{n}$ diverges, the series converges.
5. Since $\dfrac{\sqrt{n}}{n^2+1}\le \dfrac{1}{\sqrt{n}}$ and $\sum \dfrac{1}{\sqrt{n}}$ converges, the series converges.

Explanation: The skill being tested is the comparison test for determining series convergence. To apply it to sum √n /(n^2 +1), observe that the term behaves like 1/n^{3/2} for large n. Choose to compare with sum 1/n^{3/2}, verifying that √n /(n^2 +1) ≤ √n /n^2 = 1/n^{3/2} for n ≥ 1 since the denominator is larger. Since sum 1/n^{3/2} converges as a p-series with 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 is actually the opposite, making it invalid. A transferable comparison strategy is to bound the term by ignoring lower-order terms in the denominator to establish an upper bound with a known convergent series.

Question 7

1. Diverges 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 $n$.
2. Converges by comparison with $\sum \frac{1}{n}$ since $\frac{2^n}{3^n+n}<\frac{1}{n}$ for all $n$.
3. 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 $n$. (correct answer)
4. Diverges by limit comparison with $\sum \left(\frac{2}{3}\right)^n$ since $\lim_{n\to\infty}\frac{\frac{2^n}{3^n+n}}{\left(\frac{2}{3}\right)^n}=0$.
5. Converges by limit comparison with $\sum \frac{1}{n^2}$ since $\lim_{n\to\infty}\frac{\frac{2^n}{3^n+n}}{\frac{1}{n^2}}=0$.

Explanation: 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.

Question 8

1. Converges by comparison with $\sum \frac{1}{n^2}$ since $\frac{n^2}{2n^3+7}<\frac{1}{n^2}$ for large $n$.
2. Diverges by limit comparison with $\sum \frac{1}{n}$ since $\lim_{n\to\infty}\frac{\frac{n^2}{2n^3+7}}{\frac{1}{n}}=\frac{1}{2}$. (correct answer)
3. Converges by limit comparison with $\sum \frac{1}{n}$ since $\lim_{n\to\infty}\frac{\frac{n^2}{2n^3+7}}{\frac{1}{n}}=0$.
4. Diverges by comparison with $\sum \frac{1}{n^2}$ since $\frac{n^2}{2n^3+7}>\frac{1}{n^2}$ for all $n$.
5. Converges by comparison with $\sum \frac{1}{n^3}$ since $\frac{n^2}{2n^3+7}<\frac{1}{n^3}$ for large $n$.

Explanation: 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.

Question 9

1. Since $\frac{n+2}{n^2+\sqrt{n}}\le \frac{2n}{n^2}=\frac{2}{n}$ for $n\ge2$ and $\sum \frac{2}{n}$ diverges, the series converges.
2. Since $\frac{n+2}{n^2+\sqrt{n}}\ge \frac{1}{n}$ for $n\ge1$ and $\sum \frac{1}{n}$ diverges, the series diverges. (correct answer)
3. Since $\frac{n+2}{n^2+\sqrt{n}}\le \frac{1}{n^2}$ for $n\ge1$ and $\sum \frac{1}{n^2}$ converges, the series diverges.
4. Since $\frac{n+2}{n^2+\sqrt{n}}\ge \frac{1}{n^2}$ for $n\ge1$ and $\sum \frac{1}{n^2}$ converges, the series converges.
5. Since $\frac{n+2}{n^2+\sqrt{n}}\le \frac{2}{n}$ for $n\ge2$ and $\sum \frac{2}{n}$ diverges, the series diverges.

Explanation: This problem requires careful analysis of the dominant terms. For large n, the numerator behaves like n and the denominator like n², giving behavior similar to 1/n. More precisely, (n+2)/(n²+√n) > n/(n²+n²) = n/(2n²) = 1/(2n) for large n. Since ∑1/n diverges (harmonic series), and our series terms are eventually larger than a constant multiple of 1/n, our series diverges by comparison. Choice A incorrectly concludes convergence from comparison with a divergent series. When the dominant behavior suggests divergence, look for a divergent series that is smaller than yours to confirm divergence.

Question 10

1. Since $\frac{2^n}{3^n+n}\ge \left(\frac{2}{3}\right)^n$ for $n\ge1$ and $\sum \left(\frac{2}{3}\right)^n$ converges, the series diverges.
2. Since $\frac{2^n}{3^n+n}\le \left(\frac{2}{3}\right)^n$ for $n\ge1$ and $\sum \left(\frac{2}{3}\right)^n$ converges, the series converges. (correct answer)
3. Since $\frac{2^n}{3^n+n}\le \left(\frac{3}{2}\right)^n$ for $n\ge1$ and $\sum \left(\frac{3}{2}\right)^n$ diverges, the series converges.
4. Since $\frac{2^n}{3^n+n}\ge \frac{1}{n}$ for $n\ge1$ and $\sum \frac{1}{n}$ diverges, the series converges.
5. Since $\frac{2^n}{3^n+n}\le \frac{1}{n}$ for $n\ge1$ and $\sum \frac{1}{n}$ diverges, the series diverges.

Explanation: This problem requires comparing an exponential series for convergence. Since 3ⁿ + n > 3ⁿ, we have 2ⁿ/(3ⁿ+n) < 2ⁿ/3ⁿ = (2/3)ⁿ. The geometric series ∑(2/3)ⁿ converges because |2/3| < 1. By the comparison test, since our series terms are smaller than those of a convergent series, our series converges. Choice A incorrectly states the inequality as ≥ instead of ≤, which would not allow us to conclude convergence. For exponential series, identify the dominant exponential term and compare with the corresponding geometric series, remembering that ∑rⁿ converges if and only if |r| < 1.

Question 11

1. Since $\frac{n}{n^3+7n+1}\ge \frac{1}{n^2}$ for $n\ge1$ and $\sum \frac{1}{n^2}$ converges, the series converges.
2. Since $\frac{n}{n^3+7n+1}\le \frac{1}{n}$ for $n\ge1$ and $\sum \frac{1}{n}$ diverges, the series diverges.
3. 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. (correct answer)
4. Since $\frac{n}{n^3+7n+1}\ge \frac{1}{n}$ for $n\ge1$ and $\sum \frac{1}{n}$ diverges, the series converges.
5. Since $\frac{n}{n^3+7n+1}\le \frac{1}{n^3}$ for $n\ge1$ and $\sum \frac{1}{n^3}$ converges, the series diverges.

Explanation: 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.

Question 12

1. Since $\dfrac{n^2}{n^4+5n}\ge \dfrac{1}{n^2}$ and $\sum \dfrac{1}{n^2}$ converges, the series diverges.
2. Since $\dfrac{n^2}{n^4+5n}\le \dfrac{1}{n^3}$ for $n\ge1$ and $\sum \dfrac{1}{n^3}$ diverges, the series diverges.
3. Since $\dfrac{n^2}{n^4+5n}\le \dfrac{1}{n^2}$ for $n\ge1$ and $\sum \dfrac{1}{n^2}$ converges, the series converges. (correct answer)
4. Since $\dfrac{n^2}{n^4+5n}\ge \dfrac{1}{n}$ and $\sum \dfrac{1}{n}$ diverges, the series converges.
5. Since $\dfrac{n^2}{n^4+5n}\le \dfrac{1}{n}$ and $\sum \dfrac{1}{n}$ diverges, the series converges.

Explanation: The skill being tested is the comparison test for determining series convergence. To apply it to sum n^2 /(n^4 +5n), note the term behaves like 1/n^2 for large n. Choose to compare with sum 1/n^2, verifying that n^2 /(n^4 +5n) ≤ n^2 /n^4 = 1/n^2 for n ≥ 1 since the denominator is larger. Since sum 1/n^2 converges (p=2>1), the original series converges by comparison. A tempting distractor is choice E, which uses a_n ≤ 1/n with the divergent harmonic series to conclude convergence, but this is inconclusive for convergence. A transferable comparison strategy is to find an upper bound by minimizing the denominator with dominant terms for convergence conclusions.

Question 13

1. Since $\frac{\ln(n+1)}{n^2}\ge \frac{1}{n}$ and $\sum \frac{1}{n}$ diverges, the series diverges.
2. Since $\frac{\ln(n+1)}{n^2}\le \frac{1}{n}$ for $n\ge1$ and $\sum \frac{1}{n}$ diverges, the series converges.
3. Since $\frac{\ln(n+1)}{n^2}\ge \frac{1}{n^2}$ and $\sum \frac{1}{n^2}$ converges, the series diverges.
4. Since $\frac{\ln(n+1)}{n^2}\le \frac{\ln(n+1)}{n}$ and $\sum \frac{\ln(n+1)}{n}$ converges, the series converges.
5. 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. (correct answer)

Explanation: 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 $n$.

Question 14

1. Since $\dfrac{\ln(n+1)}{n^2}\ge \dfrac{1}{n}$ and $\sum \dfrac{1}{n}$ diverges, the series converges.
2. Since $\dfrac{\ln(n+1)}{n^2}\le \dfrac{1}{n}$ and $\sum \dfrac{1}{n}$ diverges, the series converges.
3. Since $\dfrac{\ln(n+1)}{n^2}\le \dfrac{n}{n^2}=\dfrac{1}{n}$ for $n\ge1$ and $\sum \dfrac{1}{n}$ diverges, the series diverges.
4. Since $\dfrac{\ln(n+1)}{n^2}\le \dfrac{1}{n^{3/2}}$ for $n\ge1$ and $\sum \dfrac{1}{n^{3/2}}$ converges, the series converges. (correct answer)
5. Since $\dfrac{\ln(n+1)}{n^2}\ge \dfrac{1}{n^{3/2}}$ and $\sum \dfrac{1}{n^{3/2}}$ converges, the series diverges.

Explanation: The skill being tested is the comparison test for determining series convergence. To apply it to sum ln(n+1)/n^2, note the term behaves like (ln n)/n^2, which is smaller than 1/n^{3/2} for large n. Choose to compare with sum 1/n^{3/2}, verifying that ln(n+1)/n^2 ≤ 1/n^{3/2} for n ≥ 1 since ln(n+1) ≤ √n holds. Since sum 1/n^{3/2} converges (p=3/2>1), the original series converges by comparison. A tempting distractor is choice B, which uses ln(n+1)/n^2 ≤ 1/n with the divergent harmonic series to conclude convergence, but this comparison is inconclusive for convergence. A transferable comparison strategy is to bound slowly growing functions like logarithms with polynomial terms to compare with p-series.

Question 15

1. Since $\dfrac{1}{\sqrt{n^3+n}}\ge \dfrac{1}{n}$ and $\sum \dfrac{1}{n}$ diverges, the series converges.
2. 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. (correct answer)
3. Since $\dfrac{1}{\sqrt{n^3+n}}\ge \dfrac{1}{n^{3/2}}$ and $\sum \dfrac{1}{n^{3/2}}$ converges, the series diverges.
4. Since $\dfrac{1}{\sqrt{n^3+n}}\le \dfrac{1}{n}$ and $\sum \dfrac{1}{n}$ diverges, the series converges.
5. Since $\dfrac{1}{\sqrt{n^3+n}}\le \dfrac{1}{\sqrt{n}}$ and $\sum \dfrac{1}{\sqrt{n}}$ converges, the series converges.

Explanation: 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.

Question 16

1. Since $a_n\ge \dfrac{3}{n^2}$ and $\sum \dfrac{3}{n^2}$ converges, $\sum a_n$ diverges.
2. Since $a_n\le \dfrac{4}{n}$ and $\sum \dfrac{4}{n}$ diverges, $\sum a_n$ converges.
3. Since $a_n\le \dfrac{4}{n^2}$ for $n\ge1$ and $\sum \dfrac{4}{n^2}$ converges, $\sum a_n$ converges. (correct answer)
4. Since $a_n\ge \dfrac{1}{n^3}$ and $\sum \dfrac{1}{n^3}$ converges, $\sum a_n$ diverges.
5. Since $a_n\le \dfrac{1}{n}$ and $\sum \dfrac{1}{n}$ diverges, $\sum a_n$ diverges.

Explanation: The skill being tested is the comparison test for determining series convergence. To apply it to sum (3n+1)/(n^3+4), identify a bounding series by noting the term behaves like 3/n^2 for large n. Choose to compare with sum 4/n^2, verifying that (3n+1)/(n^3+4) ≤ 4/n^2 for n ≥ 1 through algebraic manipulation. Since sum 4/n^2 converges as a p-series with p=2>1, the original series converges by the comparison test. A tempting distractor is choice A, which claims a_n ≥ 3/n^2 and concludes divergence, but a lower bound by a convergent series is inconclusive for divergence. A transferable comparison strategy is to approximate terms with dominant polynomials and adjust constants to establish strict inequalities with known p-series.

Question 17

1. Since $\dfrac{1}{n^2+n}\ge \dfrac{1}{n^2}$ and $\sum \dfrac{1}{n^2}$ converges, the series diverges.
2. Since $\dfrac{1}{n^2+n}\le \dfrac{1}{n^2}$ for $n\ge1$ and $\sum \dfrac{1}{n^2}$ converges, the series converges. (correct answer)
3. Since $\dfrac{1}{n^2+n}\le \dfrac{1}{n}$ and $\sum \dfrac{1}{n}$ diverges, the series converges.
4. Since $\dfrac{1}{n^2+n}\ge \dfrac{1}{n}$ and $\sum \dfrac{1}{n}$ diverges, the series converges.
5. Since $\dfrac{1}{n^2+n}\le \dfrac{1}{n^3}$ and $\sum \dfrac{1}{n^3}$ diverges, the series diverges.

Explanation: The skill being tested is the comparison test for determining series convergence. To apply it to sum 1/(n^2 + n), note the term behaves like 1/n^2 for large n. Choose to compare with sum 1/n^2, verifying that 1/(n^2 + n) ≤ 1/n^2 for n ≥ 1 since the denominator is larger. Since sum 1/n^2 converges (p=2>1), the original series converges by comparison. A tempting distractor is choice A, which claims a_n ≥ 1/n^2 and concludes divergence, but while the inequality is true, the logic is flawed as a lower bound by convergent doesn't imply divergence. A transferable comparison strategy is to add positive terms to the denominator to create an upper bound for convergence.

Question 18

1. Since $\dfrac{2+\sin n}{n^2}\ge \dfrac{3}{n^2}$ and $\sum \dfrac{3}{n^2}$ converges, the series diverges.
2. Since $\dfrac{2+\sin n}{n^2}\le \dfrac{1}{n}$ and $\sum \dfrac{1}{n}$ diverges, the series converges.
3. Since $\dfrac{2+\sin n}{n^2}\le \dfrac{1}{n^2}$ and $\sum \dfrac{1}{n^2}$ diverges, the series diverges.
4. Since $\dfrac{2+\sin n}{n^2}\le \dfrac{3}{n^2}$ and $\sum \dfrac{3}{n^2}$ converges, the series converges. (correct answer)
5. Since $\dfrac{2+\sin n}{n^2}\ge \dfrac{1}{n}$ and $\sum \dfrac{1}{n}$ diverges, the series converges.

Explanation: 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.

Question 19

1. Converges by limit comparison with $\sum \frac{1}{n^2}$ since $\lim_{n\to\infty}\frac{\frac{3n+1}{n^3+5}}{\frac{1}{n^2}}=3$. (correct answer)
2. Diverges by comparison with $\sum \frac{1}{n}$ since $\frac{3n+1}{n^3+5}>\frac{1}{n}$ for large $n$.
3. Converges by comparison with $\sum \frac{1}{n}$ since $\frac{3n+1}{n^3+5}<\frac{1}{n}$ for all $n$.
4. Diverges by limit comparison with $\sum \frac{1}{n^2}$ since $\lim_{n\to\infty}\frac{\frac{3n+1}{n^3+5}}{\frac{1}{n^2}}=0$.
5. Converges by comparison with $\sum \frac{1}{n^3}$ since $\frac{3n+1}{n^3+5}<\frac{1}{n^3}$ for large $n$.

Explanation: 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.

Question 20

1. Converges by comparison with $\sum \frac{1}{n^2}$ since $\frac{n}{n^2+\sin n}<\frac{1}{n^2}$ for all $n$.
2. Diverges by limit comparison with $\sum \frac{1}{n}$ since $\lim_{n\to\infty}\frac{\frac{n}{n^2+\sin n}}{\frac{1}{n}}=1$. (correct answer)
3. Converges by limit comparison with $\sum \frac{1}{n}$ since $\lim_{n\to\infty}\frac{\frac{n}{n^2+\sin n}}{\frac{1}{n}}=0$.
4. Converges by comparison with $\sum \frac{1}{n}$ since $\frac{n}{n^2+\sin n}<\frac{1}{n}$ for all $n$.
5. Diverges by comparison with $\sum \frac{1}{n^2}$ since $\frac{n}{n^2+\sin n}>\frac{1}{n^2}$ for large $n$.

Explanation: 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.