Alternating Series Test for Convergence - AP Calculus BC

A physics model includes $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n!}$ . Does the series converge?
Answer: Converges by the Alternating Series Test because $\frac{1}{n!}$ decreases and approaches $0$ . The skill being tested is the Alternating Series Test for convergence. Conditions: $b_n$ positive, decreasing, lim=0. $b_n$ = 1/n! decreases rapidly to 0, so converges (absolutely too). Factorials ensure quick decay. Choice E distracts since ∑ 1/n! converges (e-1), not diverges; recall e series. For exponential decay like factorials, expect absolute convergence but test still applies.
A series is $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{(n+1)\sqrt{n+1}}$ . Does it converge?
Answer: Converges by the Alternating Series Test because $\frac{1}{(n+1)\sqrt{n+1}}$ decreases and approaches $0$ . The skill being tested here is the Alternating Series Test for convergence. Conditions include $b_n$ > 0, monotonic decrease, and limit 0 for convergence. This applies to p-series-like terms with p > 1/2 here. For ∑ $(-1)^{n+1}$ \frac{1}{(n+1) sqrt{n+1}}, $b_n$ decreases to 0 like 1/ $n^{3/2}$ . A tempting distractor is choice C, claiming divergence because limit ≠ 0, but it does approach 0. A transferable strategy for alternating series is to rewrite $b_n$ in terms of powers to assess decrease and limit.
A series is $\sum_{n=1}^{\infty}(-1)^n\frac{1}{\ln(n)}$ with $n\ge2$ . Does it converge?
Answer: Converges by the Alternating Series Test because $\frac{1}{\ln n}$ decreases and approaches $0$ . The skill being tested here is the Alternating Series Test for convergence. It requires $b_n$ > 0, monotonically decreasing, and lim $b_n$ = 0, even if slowly. This allows convergence despite absolute divergence. For ∑ $(-1)^n$ \frac{1}{ln n} (n ≥ 2), $b_n$ decreases to 0, satisfying the conditions. A tempting distractor is choice C, saying it diverges because the absolute series diverges, but AST doesn't require that. A transferable strategy for alternating series is to test monotonicity by considering the derivative of the continuous version.
A computation uses $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\ln n}{n}$ . Does the series converge?
Answer: Converges by the Alternating Series Test because $\frac{\ln n}{n}$ decreases for large $n$ and approaches $0$ . 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 $b_n$ .
A series is $\sum_{n=1}^{\infty}(-1)^n\frac{1}{n^{1/3}}$ . Does it converge?
Answer: Converges by the Alternating Series Test because $\frac{1}{n^{1/3}}$ decreases and approaches $0$ . The skill being tested here is the Alternating Series Test for convergence. For p-series alternated, converges if p > 0 since lim = 0 and decreasing. Absolute convergence needs p > 1. For ∑ $(-1)^n$ \frac{1}{n^{1/3}}, $b_n$ decreases to 0, converges. A tempting distractor is choice B, saying diverges for p < 1, but that's for absolute series. A transferable strategy for alternating series is to recall AST works for 0 < p ≤ 1 unlike p-test.
A series is $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n+\ln n}$ for $n\ge2$ . Does it converge?
Answer: Converges by the Alternating Series Test because $\frac{1}{n+\ln n}$ decreases and approaches $0$ . The skill being tested here is the Alternating Series Test for convergence. $b_n$ must be positive, decreasing, approach 0; logs grow slowly but still work. Compare to 1/n. For ∑ $(-1)^{n+1}$ \frac{1}{n + ln n} (n ≥ 2), $b_n$ decreases to 0, converges. A tempting distractor is choice A, saying divergence because denominator increases, but that's why it decreases. A transferable strategy for alternating series is to ignore slow-growing terms like ln n for large-n behavior.
A series is $\sum_{n=1}^{\infty}(-1)^n\frac{n}{n+2}$ . Does it converge?
Answer: Diverges because $\lim_{n\to\infty}\frac{n}{n+2}\neq 0$ . The skill being tested here is the Alternating Series Test for convergence. For convergence, $b_n$ must be positive, decreasing, and lim $b_n$ = 0; otherwise, it fails. The limit condition is a quick first check. Here, for ∑ $(-1)^n$ \frac{n}{n+2}, $b_n$ approaches 1 ≠ 0, so it diverges. A tempting distractor is choice A, assuming convergence because $b_n$ decreases to 0, but the limit is 1. A transferable strategy for alternating series is to compute the limit of $b_n$ before checking other conditions.
A series for alternating work is $\sum_{n=1}^{\infty}(-1)^n\frac{\arctan(n)}{n}$ . Does it converge?
Answer: Converges by the Alternating Series Test because $\frac{\arctan(n)}{n}$ decreases for large $n$ and approaches $0$ . 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.
A power-series remainder uses $\sum_{n=1}^{\infty}(-1)^n\frac{1}{n^3}$ . Does the series converge?
Answer: Converges by the Alternating Series Test because $\frac{1}{n^3}$ decreases and approaches $0$ . 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.
A Fourier-like term is $\sum_{n=1}^{\infty}(-1)^{n}\frac{2n+1}{n}$ . Does it converge?
Answer: Diverges because $\lim_{n\to\infty}\frac{2n+1}{n}\neq 0$ . 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.
A series for a measurement error is $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{n^3+1}$ . Does it converge?
Answer: Converges by the Alternating Series Test because $\frac{n}{n^3+1}$ decreases and approaches $0$ . 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.
A series is $\sum_{n=1}^{\infty}(-1)^n\frac{n+2}{n^2}$ . Does it converge?
Answer: Converges by the Alternating Series Test because $\frac{n+2}{n^2}$ decreases and approaches $0$ . 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 ∑ $(-1)^n$ \frac{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.
A signal correction uses $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$ . Does this series converge?
Answer: Converges by the Alternating Series Test because $\frac{1}{n}$ decreases and approaches $0$ . The skill being tested is the Alternating Series Test for convergence. This test applies to series of the form ∑ $(-1)^{n+1}$ $b_n$ where $b_n$ > 0, the sequence $b_n$ is decreasing, and lim $b_n$ = 0, guaranteeing convergence. For the series ∑ $(-1)^{n+1}$ (1/n), $b_n$ = 1/n is positive, decreasing for all n, and approaches 0 as n goes to infinity. Thus, the conditions are satisfied, and the series converges conditionally since the harmonic series diverges. A tempting distractor is choice D, which claims divergence because lim (1/n) ≠ 0, but this ignores the alternating signs that enable convergence via the test. Always check both the limit and monotonicity conditions when applying the Alternating Series Test to avoid missing conditional convergence.
An error estimate uses $\sum_{n=1}^{\infty}(-1)^{n}\frac{n}{n+1}$ . Does the series converge?
Answer: Diverges because $\lim_{n\to\infty}\frac{n}{n+1}\neq 0$ . 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.
A convergence check involves $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{\sqrt[n]{n}}$ . Does it converge?
Answer: Diverges because $\lim_{n\to\infty}\frac{1}{\sqrt[n]{n}}\neq 0$ . The skill being tested is the Alternating Series Test for convergence. Requires $b_n$ decreasing to 0. Here, $b_n$ = 1/ $n^{1/n}$ → 1 ≠ 0 since $n^{1/n}$ →1, so diverges. Terms alternate around 1. Choice A distracts by claiming decrease to 0, but limit is 1; compute special limits carefully. Know that $n^{1/n}$ approaches 1, impacting root-related series.
A sequence of adjustments is $\sum_{n=1}^{\infty}(-1)^n\frac{1}{n+\sin n}$ . Does it converge?
Answer: Diverges because $\frac{1}{n+\sin n}$ is not monotone decreasing. The skill being tested here is the Alternating Series Test for convergence. The test demands strict monotonicity in $b_n$ , positivity, and limit 0. Lack of monotonicity can cause unbounded partial sums. For ∑ $(-1)^n$ \frac{1}{n + sin n}, $b_n$ oscillates due to sin n, not monotone, so diverges. A tempting distractor is choice A, assuming it decreases to 0, but oscillation prevents monotonicity. A transferable strategy for alternating series is to check $b_{n+1}$ - $b_n$ < 0 for consecutive terms.
A series for corrections is $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n(\ln n)^2}$ for $n\ge2$ . Does it converge?
Answer: Converges by the Alternating Series Test because $\frac{1}{n(\ln n)^2}$ decreases and approaches $0$ . 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 ∑ $(-1)^{n+1}$ \frac{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.
A model for alternating drift uses $\sum_{n=1}^{\infty}(-1)^n\frac{n}{n+1}$ . Does the series converge?
Answer: Diverges because $\frac{n}{n+1}\not\to 0$ . This question requires applying the alternating series test for convergence. The series $\sum_{n=1}^{\infty}(-1)^n\frac{n}{n+1}$ has alternating signs, so we check if $a_n = \frac{n}{n+1}$ satisfies the test conditions. First, we find $\lim_{n\to\infty} \frac{n}{n+1} = \lim_{n\to\infty} \frac{1}{1+1/n} = 1 \neq 0$ . Since the terms don't approach zero, the series diverges by the divergence test (also called the nth term test). The alternating series test cannot even be applied because its first requirement—that terms approach zero—is violated. Choice A incorrectly focuses only on the alternating and decreasing properties without checking the limit. Remember: for any series to converge, alternating or not, the terms must approach zero.
A calibration update is $\sum_{n=1}^{\infty}(-1)^n\frac{1}{\ln(n+1)}$ . Does the series converge?
Answer: Converges because $\frac{1}{\ln(n+1)}\to 0$ and is decreasing. This question tests the alternating series test on $\sum_{n=1}^{\infty}(-1)^n\frac{1}{\ln(n+1)}$ . Since $\ln(n+1)$ increases without bound, we have $\lim_{n\to\infty} \frac{1}{\ln(n+1)} = 0$ . To verify decreasing behavior, note that if $f(x) = \frac{1}{\ln(x+1)}$ , then $f'(x) = -\frac{1}{(x+1)[ln(x+1)]^2} < 0$ for all $x > 0$ . Thus $a_n = \frac{1}{\ln(n+1)}$ is decreasing for all $n \geq 1$ . Both conditions are met, so the series converges. Choice D incorrectly suggests the terms decrease "too slowly"—the rate of decrease doesn't matter as long as the terms are decreasing and approach zero. The alternating series test is remarkably forgiving: any decreasing sequence approaching zero yields convergence.
An algorithm sums $\sum_{n=1}^{\infty}(-1)^n\frac{\sqrt{n}}{n+1}$ . Does the series converge?
Answer: Converges by the Alternating Series Test because $\frac{\sqrt{n}}{n+1}$ decreases and approaches $0$ . 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.

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