AP Calculus BC Quiz: Alternating Series Test For Convergence

What this quiz covers

This quiz focuses on Alternating Series Test 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. Diverges because $\frac{n}{n^2+1}$ is not decreasing.
2. Converges absolutely because $\sum \frac{n}{n^2+1}$ converges.
3. Converges by the Alternating Series Test because $\frac{n}{n^2+1}$ decreases and approaches $0$. (correct answer)
4. Diverges because $\lim_{n\to\infty}\frac{n}{n^2+1}=1$.
5. Converges because any alternating rational function converges.

Explanation: The skill being tested is the Alternating Series Test for convergence. The test applies when b_n > 0, is eventually decreasing, and limits to 0. For b_n = n/(n^2+1), it approaches 0 like 1/n, and is decreasing for n ≥ 1 as its derivative is negative. Thus, the series converges by the test. Choice D is tempting but wrong because lim n/(n^2+1) = 0, not 1; miscalculating the limit leads to errors. A key strategy is to analyze the asymptotic behavior of b_n to confirm it decreases and vanishes at infinity.

Question 2

1. Converges by the Alternating Series Test because $\frac{\sin(1/n)}{n}$ decreases for large $n$ and approaches $0$. (correct answer)
2. Diverges because $\sin(1/n)$ oscillates.
3. Diverges because $\lim_{n\to\infty}\frac{\sin(1/n)}{n}\neq 0$.
4. Converges because $\sin(1/n)$ is negative for large $n$.
5. Converges absolutely because $\sum \frac{\sin(1/n)}{n}$ converges.

Explanation: The skill being tested here is the Alternating Series Test for convergence. The conditions are b_n positive, decreasing (eventually), and limiting to 0 for convergence. Eventual monotonicity suffices for large n. For ∑ (-1)^{n+1} rac{sin(1/n)}{n}, b_n ≈ 1/n^2 for large n, decreasing to 0, so it converges. A tempting distractor is choice B, claiming divergence due to oscillation in sin, but sin(1/n) is positive and the whole b_n decreases. A transferable strategy for alternating series is to use approximations like Taylor expansions for complex b_n.

Question 3

1. Converges by the Alternating Series Test because $\frac{1}{n+(-1)^n}$ decreases and approaches $0$.
2. Diverges because $\frac{1}{n+(-1)^n}$ is not monotone decreasing. (correct answer)
3. Converges absolutely because $\sum \frac{1}{n+(-1)^n}$ converges.
4. Diverges because $\lim_{n\to\infty}\frac{1}{n+(-1)^n}\neq 0$.
5. Converges because the denominator alternates.

Explanation: The skill being tested here is the Alternating Series Test for convergence. For ∑ (-1)^n b_n to converge, b_n must be positive, the sequence must be monotonically decreasing, and lim b_n = 0. Monotonicity is crucial to prevent oscillations that could cause divergence. In this case, b_n = rac{1}{n + (-1)^n} oscillates and is not monotonically decreasing, so the series diverges. A tempting distractor is choice A, which assumes it decreases to 0, but the oscillation violates monotonicity. A transferable strategy for alternating series is to plot or compute a few terms to check for strict monotonicity in b_n.

Question 4

1. Converges by the Alternating Series Test because $\frac{1}{1+\sin^2 n}$ decreases and approaches $0$.
2. Diverges because $\lim_{n\to\infty}\frac{1}{1+\sin^2 n}\neq 0$. (correct answer)
3. Converges because $\frac{1}{1+\sin^2 n}$ is bounded.
4. Converges absolutely because $\sum \frac{1}{1+\sin^2 n}$ converges.
5. Converges because $\sin^2 n$ is periodic.

Explanation: The skill being tested is the Alternating Series Test for convergence. Test requires lim b_n=0 among others. Here, b_n = 1/(1+sin^2 n) doesn't →0 as it oscillates between 1/2 and 1 densely. Thus, diverges by nth-term test. Choice A tempts but fails on limit condition due to oscillation. Be cautious with oscillatory terms; ensure the limit exists and is zero.

Question 5

1. Diverges because $\frac{1}{\ln n}$ is increasing.
2. Converges by the Alternating Series Test because $\frac{1}{\ln n}$ decreases and approaches $0$. (correct answer)
3. Converges absolutely because $\sum \frac{1}{\ln n}$ converges.
4. Diverges because $\sum \frac{1}{\ln n}$ diverges, so the alternating series diverges.
5. Converges because $\lim_{n\to\infty}\frac{1}{\ln n}=1$.

Explanation: The skill being tested is the Alternating Series Test for convergence. This test requires an alternating series ∑ (-1)^n b_n with b_n positive, decreasing to 0 in the limit. Here, b_n = 1/ln n for n ≥ 2 is positive, decreases because ln n increases slower than any positive power, and lim (1/ln n) = 0. Therefore, the series converges by the test. Choice D is tempting but fails because divergence of the absolute series does not imply divergence of the alternating one; the test specifically allows conditional convergence. Remember to verify the decreasing condition for n beyond a certain point, as initial terms may not strictly decrease but the tail determines convergence.

Question 6

1. Diverges because $\frac{n}{n^3+1}$ does not approach $0$.
2. Converges because $\sum \frac{n}{n^3+1}$ diverges.
3. Converges by the Alternating Series Test because $\frac{n}{n^3+1}$ decreases and approaches $0$. (correct answer)
4. Diverges because $\frac{n}{n^3+1}$ is increasing.
5. Diverges because alternating series require $b_n$ to increase.

Explanation: The skill being tested is the Alternating Series Test for convergence. The conditions are b_n > 0, eventually decreasing, and lim b_n = 0. For b_n = n/(n^3+1) ≈ 1/n^2, it limits to 0 and decreases for n ≥ 1 based on derivative analysis. Thus, the series converges. Choice A distracts by claiming it doesn't approach 0, but it does; always compute limits carefully. Use approximation for large n to quickly check both conditions in alternating series.

Question 7

1. Converges by the Alternating Series Test because $\frac{1}{n(\ln n)^2}$ decreases and approaches $0$. (correct answer)
2. Diverges because $\frac{1}{n(\ln n)^2}$ increases.
3. Diverges because $\lim_{n\to\infty}\frac{1}{n(\ln n)^2}=1$.
4. Converges because $\sum \frac{1}{n(\ln n)^2}$ diverges.
5. Diverges because alternating series require $b_n$ to be constant.

Explanation: The skill being tested here is the Alternating Series Test for convergence. The test checks if b_n is positive, decreasing, and approaches 0, ensuring bounded partial sums. This is useful for series with slow-decaying terms. For ∑ (-1)^{n+1} rac{1}{n (ln n)^2} (n ≥ 2), b_n decreases to 0, so it converges. A tempting distractor is choice D, claiming convergence because the absolute diverges, but that's irrelevant to AST. A transferable strategy for alternating series is to use integral tests on |b_n| for additional insights, but not as a requirement.

Question 8

1. Converges by the Alternating Series Test because $\frac{\arctan(n)}{n}$ decreases for large $n$ and approaches $0$. (correct answer)
2. Diverges because $\arctan(n)$ does not approach $0$.
3. Converges absolutely because $\sum \frac{\arctan(n)}{n}$ converges.
4. Diverges because $\frac{\arctan(n)}{n}$ is increasing.
5. Diverges because alternating series require $b_n$ to be constant.

Explanation: The skill being tested is the Alternating Series Test for convergence. Test needs b_n eventually decreasing to 0. b_n = arctan(n)/n → 0 since arctan bounds to π/2, and decreases after initial terms. Thus, converges. Choice B is tempting but wrong as arctan(n) → π/2, but divided by n →0. Examine bounded numerators over growing denominators for limit zero in such series.

Question 9

1. Converges by the Alternating Series Test because $\frac{2n+1}{n}$ decreases to $0$.
2. Diverges because $\lim_{n\to\infty}\frac{2n+1}{n}\neq 0$. (correct answer)
3. Converges because the terms alternate in sign.
4. Converges absolutely because $\sum \frac{2n+1}{n}$ converges.
5. Diverges because $\frac{2n+1}{n}$ is decreasing.

Explanation: The skill being tested is the Alternating Series Test for convergence. The test requires $b_n > 0$, decreasing, and approaching 0. Here, $b_n = \frac{2n+1}{n} = 2 + \frac{1}{n}$ approaches $2 \neq 0$, so terms do not go to zero, and the series diverges. The alternating nature cannot overcome non-vanishing terms. Choice A is a distractor that falsely claims it decreases to 0, but it decreases to 2, violating the limit condition. Prioritize checking the limit of $b_n$ before assessing monotonicity in alternating series analysis.

Question 10

1. Diverges because $\frac{n+2}{n^2}$ is increasing.
2. Converges by the Alternating Series Test because $\frac{n+2}{n^2}$ decreases and approaches $0$. (correct answer)
3. Diverges because $\lim_{n\to\infty}\frac{n+2}{n^2}=1$.
4. Converges absolutely because $\sum \frac{n+2}{n^2}$ diverges.
5. Diverges because alternating series require $b_n$ to be increasing.

Explanation: The skill being tested here is the Alternating Series Test for convergence. To apply it, confirm b_n > 0, b_n is decreasing (at least eventually), and lim b_n = 0. These ensure the remainders decrease appropriately. For ∑ (-1)^n rac{n+2}{n^2}, b_n approaches 0 and decreases for n ≥ 1, satisfying the test. A tempting distractor is choice A, saying it diverges because b_n increases, but actually it decreases like 1/n. A transferable strategy for alternating series is to simplify b_n asymptotically to check the limit and monotonicity.

Question 11

1. Converges by the Alternating Series Test because $\frac{1}{n^3}$ decreases and approaches $0$. (correct answer)
2. Diverges because $\sum \frac{1}{n^3}$ diverges.
3. Diverges because $\lim_{n\to\infty}\frac{1}{n^3}\neq 0$.
4. Converges because $\frac{1}{n^3}$ is increasing.
5. Diverges because it is an alternating harmonic series.

Explanation: The skill being tested is the Alternating Series Test for convergence. It demands b_n positive, decreasing, and approaching 0. Here, b_n = 1/n^3 is clearly positive, decreasing, and limits to 0, so converges (and absolutely as p=3>1). The test confirms convergence regardless. Choice B is a distractor since ∑ 1/n^3 actually converges, not diverges; check p-value correctly. When b_n is a p-series term, recall p>1 for absolute but test works for p>0 with lim=0.

Question 12

1. Converges by the Alternating Series Test because $\frac{n}{2^n}$ decreases for large $n$ and approaches $0$. (correct answer)
2. Diverges because $\frac{n}{2^n}$ increases without bound.
3. Diverges because $\lim_{n\to\infty}\frac{n}{2^n}=1$.
4. Converges because the terms do not alternate.
5. Diverges because $\sum \frac{n}{2^n}$ diverges.

Explanation: The skill being tested is the Alternating Series Test for convergence. It demands eventual decrease of b_n to 0. b_n = n/2^n →0 by ratio test idea, and decreases after n=2. Thus, converges. Choice B is a distractor as it decreases eventually, not increases; check sequences numerically. For polynomial over exponential, expect eventual decrease and convergence.

Question 13

1. Diverges because $\sum \frac{1}{\sqrt{n}}$ diverges, so the alternating series diverges.
2. Converges because $\frac{1}{\sqrt{n}}$ decreases and approaches $0$. (correct answer)
3. Diverges because $\frac{1}{\sqrt{n}}$ is not decreasing for all $n$.
4. Converges because the ratio test limit equals $1/\sqrt{n}$.
5. Diverges because alternating signs prevent convergence.

Explanation: This problem tests the alternating series test for convergence on $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{\sqrt{n}}$. For the alternating series test, we need $a_n = \frac{1}{\sqrt{n}}$ to be eventually decreasing and approach zero. Since $\frac{d}{dn}\left(\frac{1}{\sqrt{n}}\right) = -\frac{1}{2n^{3/2}} < 0$ for all $n > 0$, the function is decreasing, and clearly $\lim_{n\to\infty} \frac{1}{\sqrt{n}} = 0$. Both conditions are satisfied, so the series converges. Choice A incorrectly assumes that divergence of the non-alternating series implies divergence of the alternating series, but alternating series can converge even when their non-alternating counterparts diverge. The key insight: alternating series test provides conditional convergence when absolute convergence fails.

Question 14

1. Converges because $\frac{1}{n+5}$ decreases and approaches $0$. (correct answer)
2. Diverges because shifting by $5$ changes convergence.
3. Diverges because $\frac{1}{n+5}$ is not decreasing.
4. Converges because the series is geometric with ratio $-1$.
5. Diverges because $\sum \frac{1}{n+5}$ diverges, so the alternating one diverges.

Explanation: This question examines the alternating series test for $\sum_{n=1}^{\infty}(-1)^n\frac{1}{n+5}$. The terms $a_n = \frac{1}{n+5}$ clearly decrease since $\frac{1}{n+6} < \frac{1}{n+5}$, and $\lim_{n\to\infty} \frac{1}{n+5} = 0$. Both conditions of the alternating series test are satisfied, so the series converges. Choice E incorrectly assumes that because $\sum \frac{1}{n+5}$ diverges (it's essentially the harmonic series shifted by 5), the alternating version must also diverge, but this reasoning is flawed. Choice B wrongly suggests that shifting indices affects convergence—it doesn't. The alternating series test works regardless of index shifts, as long as the conditions on the terms are met.

Question 15

1. Diverges because $\frac{1}{n^2+5n}$ is not positive.
2. Converges by the Alternating Series Test because $\frac{1}{n^2+5n}$ decreases and approaches $0$. (correct answer)
3. Diverges because $\lim_{n\to\infty}\frac{1}{n^2+5n}=1$.
4. Diverges because $\sum \frac{1}{n^2+5n}$ diverges.
5. Converges because $\frac{1}{n^2+5n}$ is increasing.

Explanation: The skill being tested here is the Alternating Series Test for convergence. The test needs b_n positive, decreasing monotonically, and limiting to 0. Partial fraction decomposition can help verify. For ∑ (-1)^{n+1} rac{1}{n^2 + 5n}, b_n = 1/(n(n+5)) decreases to 0, converges. A tempting distractor is choice D, saying divergence because absolute diverges, but AST allows that. A transferable strategy for alternating series is to decompose rational b_n to assess decrease.

Question 16

1. Converges by the Alternating Series Test because $\frac{\sqrt{n}}{n+1}$ decreases and approaches $0$. (correct answer)
2. Diverges because $\frac{\sqrt{n}}{n+1}$ approaches $1$.
3. Converges absolutely because $\sum \frac{\sqrt{n}}{n+1}$ converges.
4. Diverges because $\frac{\sqrt{n}}{n+1}$ is increasing for all $n$.
5. Converges because $\sum \frac{\sqrt{n}}{n+1}$ diverges.

Explanation: The skill being tested is the Alternating Series Test for convergence. The test requires eventual decrease of b_n to 0. For b_n = √n/(n+1) ≈ 1/√n, it limits to 0 and decreases for n ≥ 1 as per derivative. Hence, converges conditionally. Choice B is tempting but false since it approaches 0, not 1; simplify expressions asymptotically. For rational b_n, divide numerator and denominator by highest power to check limits and behavior.

Question 17

1. Converges by the Alternating Series Test because $\frac{\ln n}{n}$ decreases for large $n$ and approaches $0$. (correct answer)
2. Diverges because $\frac{\ln n}{n}$ increases without bound.
3. Converges absolutely because $\sum \frac{\ln n}{n}$ converges.
4. Diverges because $\sum \frac{\ln n}{n}$ diverges, so the alternating series diverges.
5. Diverges because $\lim_{n\to\infty}\frac{\ln n}{n}\neq 0$.

Explanation: The skill being tested is the Alternating Series Test for convergence. This involves checking if b_n is positive, eventually decreasing, and lim b_n = 0. For b_n = (ln n)/n, it limits to 0 by L'Hôpital's rule and decreases for n ≥ 3 after an initial increase. Therefore, the series converges conditionally since the absolute version diverges by integral test. Choice E distracts by claiming lim (ln n)/n ≠ 0, but it does equal 0, emphasizing careful limit evaluation. Always use calculus tools like derivatives to verify the decreasing condition for non-obvious b_n.

Question 18

1. Converges by the Alternating Series Test because $\frac{\ln(n+1)}{\sqrt{n}}$ decreases for large $n$ and approaches $0$. (correct answer)
2. Diverges because $\ln(n+1)$ increases.
3. Converges absolutely because $\sum \frac{\ln(n+1)}{\sqrt{n}}$ converges.
4. Diverges because $\lim_{n\to\infty}\frac{\ln(n+1)}{\sqrt{n}}\neq 0$.
5. Diverges because $\frac{\ln(n+1)}{\sqrt{n}}$ is not positive.

Explanation: The skill being tested is the Alternating Series Test for convergence. Requires b_n eventually decreasing to 0. b_n = ln(n+1)/√n →0 and decreases after a maximum around n=5. Thus, converges. Choice D is tempting but lim=0, not ≠0; use L'Hôpital for logs over powers. For log over root, confirm eventual decrease via derivative for AST application.

Question 19

1. Diverges because $\frac{1}{n\ln(n+1)}$ does not approach $0$.
2. Converges by the Alternating Series Test because $\frac{1}{n\ln(n+1)}$ decreases and approaches $0$. (correct answer)
3. Diverges because $\sum \frac{1}{n\ln(n+1)}$ diverges, so the alternating series diverges.
4. Converges because $\frac{1}{n\ln(n+1)}$ is increasing.
5. Converges absolutely because $\sum \frac{1}{n\ln(n+1)}$ converges.

Explanation: The skill being tested is the Alternating Series Test for convergence. It applies if b_n > 0, decreases, and lim=0. For b_n = 1/(n ln(n+1)) ≈ 1/(n ln n), it limits to 0 and decreases for n ≥ 1. Thus, converges conditionally. Choice C tempts but errs as absolute divergence doesn't force alternating divergence; test allows it. Use integral test analogies to confirm decrease and limit for logarithmic terms.

Question 20

1. Converges by the Alternating Series Test because $\frac{n}{n+1}$ decreases to $0$.
2. Diverges because $\lim_{n\to\infty}\frac{n}{n+1}\neq 0$. (correct answer)
3. Converges because the terms alternate in sign.
4. Converges absolutely because $\sum \frac{n}{n+1}$ converges.
5. Diverges because $\frac{n}{n+1}$ is not positive.

Explanation: The skill being tested is the Alternating Series Test for convergence. For ∑ (-1)^n b_n, the test demands b_n > 0, b_n decreasing, and lim b_n = 0. In this case, b_n = n/(n+1) approaches 1, not 0, so the test fails and the series diverges by the nth-term test. No other convergence test is needed since terms do not go to zero. Choice A tempts by claiming convergence, but it overlooks that lim n/(n+1) = 1 ≠ 0, violating a key condition. A useful strategy is to first check if terms approach zero before applying more advanced tests like the Alternating Series Test.