AP Calculus BC Quiz: Alternating Series Test For Convergence
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Alternating Series Test For ConvergenceQuestion 1 of 20

A series is ∑n=1∞(−1)n+11n+ln⁡n\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n+\ln n} for n≥2n\ge2. Does it converge?

Diverges because n+ln⁡nn+\ln n increases.
Converges by the Alternating Series Test because 1n+ln⁡n\frac{1}{n+\ln n} decreases and approaches 00.
Diverges because lim⁡n→∞1n+ln⁡n=1\lim_{n\to\infty}\frac{1}{n+\ln n}=1.
Converges absolutely because ∑1n+ln⁡n\sum \frac{1}{n+\ln n} converges.
Diverges because the terms do not alternate.
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AP Calculus BC Quiz

AP Calculus BC Quiz: Alternating Series Test For Convergence

Practice Alternating Series Test For Convergence in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

All questions

Question 1

A series is ∑n=1∞(−1)n+11n+ln⁡n\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n+\ln n} for n≥2n\ge2. Does it converge?

  1. Diverges because n+ln⁡nn+\ln n increases.
  2. Converges by the Alternating Series Test because 1n+ln⁡n\frac{1}{n+\ln n} decreases and approaches 00. (correct answer)
  3. Diverges because lim⁡n→∞1n+ln⁡n=1\lim_{n\to\infty}\frac{1}{n+\ln n}=1.
  4. Converges absolutely because ∑1n+ln⁡n\sum \frac{1}{n+\ln n} converges.
  5. Diverges because the terms do not alternate.
Explanation: 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.

Question 2

A series is defined by ∑n=1∞(−1)n1n+(−1)n\sum_{n=1}^{\infty}(-1)^n\frac{1}{n+(-1)^n}. Does it converge?

  1. Converges by the Alternating Series Test because 1n+(−1)n\frac{1}{n+(-1)^n} decreases and approaches 00.
  2. Diverges because 1n+(−1)n\frac{1}{n+(-1)^n} is not monotone decreasing. (correct answer)
  3. Converges absolutely because ∑1n+(−1)n\sum \frac{1}{n+(-1)^n} converges.
  4. Diverges because lim⁡n→∞1n+(−1)n≠0\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 = \frac{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 3

A series is ∑n=1∞(−1)nn+2n2\sum_{n=1}^{\infty}(-1)^n\frac{n+2}{n^2}. Does it converge?

  1. Diverges because n+2n2\frac{n+2}{n^2} is increasing.
  2. Converges by the Alternating Series Test because n+2n2\frac{n+2}{n^2} decreases and approaches 00. (correct answer)
  3. Diverges because lim⁡n→∞n+2n2=1\lim_{n\to\infty}\frac{n+2}{n^2}=1.
  4. Converges absolutely because ∑n+2n2\sum \frac{n+2}{n^2} diverges.
  5. Diverges because alternating series require bnb_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 \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.

Question 4

A series is ∑n=1∞(−1)n1n+(−1)n+1\sum_{n=1}^{\infty}(-1)^n\frac{1}{n+(-1)^{n+1}}. Does it converge?

  1. Converges by the Alternating Series Test because 1n+(−1)n+1\frac{1}{n+(-1)^{n+1}} decreases and approaches 00.
  2. Diverges because 1n+(−1)n+1\frac{1}{n+(-1)^{n+1}} is not monotone decreasing. (correct answer)
  3. Converges absolutely because ∑1n+(−1)n+1\sum \frac{1}{n+(-1)^{n+1}} converges.
  4. Diverges because lim⁡n→∞1n+(−1)n+1≠0\lim_{n\to\infty}\frac{1}{n+(-1)^{n+1}}\neq 0.
  5. Converges because the denominator alternates.
Explanation: The skill being tested here is the Alternating Series Test for convergence. Monotonicity fails if denominator alternates significantly. Check consecutive terms. For ∑ (-1)^n \frac{1}{n + (-1)^{n+1}}, b_n not monotone due to alternating denominator, diverges. A tempting distractor is choice A, assuming decrease to 0, but not monotone. A transferable strategy for alternating series is to compute b_n for even and odd n separately.

Question 5

A series for oscillations is ∑n=1∞(−1)n1n2+n\sum_{n=1}^{\infty}(-1)^n\frac{1}{n^2+n}. Does it converge?

  1. Diverges because 1n2+n\frac{1}{n^2+n} does not approach 00.
  2. Converges by the Alternating Series Test because 1n2+n\frac{1}{n^2+n} decreases and approaches 00. (correct answer)
  3. Diverges because ∑1n2+n\sum \frac{1}{n^2+n} diverges.
  4. Converges because 1n2+n\frac{1}{n^2+n} is increasing.
  5. Diverges because alternating series require bnb_n to be nonpositive.
Explanation: The skill being tested here is the Alternating Series Test for convergence. This test states that an alternating series ∑ (-1)^n b_n converges if b_n > 0, b_n is decreasing for all n, and lim b_n = 0. The conditions ensure the partial sums are bounded and converge. For ∑ (-1)^n \frac{1}{n^2 + n}, b_n = \frac{1}{n^2 + n} is positive, decreasing, and approaches 0, so it converges. A tempting distractor is choice C, which says it diverges because the absolute series diverges, but the AST does not require absolute convergence. A transferable strategy for alternating series is to verify monotonicity by checking if b_{n+1} < b_n for large n.

Question 6

A series is ∑n=1∞(−1)n1n−n−1\sum_{n=1}^{\infty}(-1)^n\frac{1}{\sqrt{n}-\sqrt{n-1}}. Does it converge?

  1. Converges by the Alternating Series Test because 1n−n−1\frac{1}{\sqrt{n}-\sqrt{n-1}} decreases to 00.
  2. Diverges because lim⁡n→∞1n−n−1≠0\lim_{n\to\infty}\frac{1}{\sqrt{n}-\sqrt{n-1}}\neq 0. (correct answer)
  3. Converges because n−n−1\sqrt{n}-\sqrt{n-1} approaches 11.
  4. Converges absolutely because ∑1n−n−1\sum \frac{1}{\sqrt{n}-\sqrt{n-1}} converges.
  5. Diverges because the denominator is decreasing.
Explanation: The skill being tested here is the Alternating Series Test for convergence. Key conditions are b_n > 0, decreasing, and lim b_n = 0; if limit fails, diverge. Rationalize to find limits. For ∑ (-1)^n \frac{1}{sqrt{n} - sqrt{n-1}}, b_n ≈ sqrt{n} / 2 → ∞ ≠ 0, so diverges. A tempting distractor is choice A, claiming convergence because it decreases to 0, but limit is infinity. A transferable strategy for alternating series is to simplify expressions like differences via conjugation.

Question 7

A series is ∑n=1∞(−1)n+1ln⁡(n)n\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\ln(n)}{n} for n≥2n\ge2. Does it converge?

  1. Diverges because ln⁡nn\frac{\ln n}{n} does not approach 00.
  2. Converges by the Alternating Series Test because ln⁡nn\frac{\ln n}{n} decreases for large nn and approaches 00. (correct answer)
  3. Converges absolutely because ∑ln⁡nn\sum \frac{\ln n}{n} converges.
  4. Diverges because ∑ln⁡nn\sum \frac{\ln n}{n} diverges, so the alternating series diverges.
  5. Diverges because ln⁡nn\frac{\ln n}{n} is increasing.
Explanation: The skill being tested here is the Alternating Series Test for convergence. Conditions include eventual decrease and lim = 0; logs require derivative check. It's conditionally convergent. For ∑ (-1)^{n+1} \frac{ln n}{n} (n ≥ 2), b_n decreases for n > e to 0, converges. A tempting distractor is choice D, linking to absolute divergence implying alternating divergence, but no. A transferable strategy for alternating series is to use L'Hôpital for limits involving logs.

Question 8

A series is ∑n=1∞(−1)n+11n(n+1)\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n(n+1)}. Does it converge?

  1. Diverges because 1n(n+1)\frac{1}{n(n+1)} does not approach 00.
  2. Converges by the Alternating Series Test because 1n(n+1)\frac{1}{n(n+1)} decreases and approaches 00. (correct answer)
  3. Diverges because ∑1n(n+1)\sum \frac{1}{n(n+1)} diverges.
  4. Converges because 1n(n+1)\frac{1}{n(n+1)} is increasing.
  5. Diverges because alternating series require bnb_n to be increasing.
Explanation: The skill being tested here is the Alternating Series Test for convergence. b_n positive, decreasing to 0; telescoping via partial fractions. Confirms decrease. For ∑ (-1)^{n+1} \frac{1}{n(n+1)}, b_n = 1/n - 1/(n+1) decreases to 0, converges. A tempting distractor is choice C, saying diverges because absolute diverges, but AST applies regardless. A transferable strategy for alternating series is to use telescoping to verify conditions explicitly.

Question 9

A damping term is ∑n=1∞(−1)nnn2+1\sum_{n=1}^{\infty}(-1)^n\frac{n}{n^2+1}. Does the series converge?

  1. Diverges because nn2+1\frac{n}{n^2+1} is not decreasing.
  2. Converges absolutely because ∑nn2+1\sum \frac{n}{n^2+1} converges.
  3. Converges by the Alternating Series Test because nn2+1\frac{n}{n^2+1} decreases and approaches 00. (correct answer)
  4. Diverges because lim⁡n→∞nn2+1=1\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/(n2+1n^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/(n2+1n^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 10

A series used in simulation is ∑n=1∞(−1)n+1sin⁡(1/n)n\sum_{n=1}^{\infty}(-1)^{n+1}\frac{\sin(1/n)}{n}. Does it converge?

  1. Converges by the Alternating Series Test because sin⁡(1/n)n\frac{\sin(1/n)}{n} decreases for large nn and approaches 00. (correct answer)
  2. Diverges because sin⁡(1/n)\sin(1/n) oscillates.
  3. Diverges because lim⁡n→∞sin⁡(1/n)n≠0\lim_{n\to\infty}\frac{\sin(1/n)}{n}\neq 0.
  4. Converges because sin⁡(1/n)\sin(1/n) is negative for large nn.
  5. Converges absolutely because ∑sin⁡(1/n)n\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} \frac{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 11

A series appears as ∑n=1∞(−1)n11+sin⁡2n\sum_{n=1}^{\infty}(-1)^n\frac{1}{1+\sin^2 n}. Does it converge?

  1. Converges by the Alternating Series Test because 11+sin⁡2n\frac{1}{1+\sin^2 n} decreases and approaches 00.
  2. Diverges because lim⁡n→∞11+sin⁡2n≠0\lim_{n\to\infty}\frac{1}{1+\sin^2 n}\neq 0. (correct answer)
  3. Converges because 11+sin⁡2n\frac{1}{1+\sin^2 n} is bounded.
  4. Converges absolutely because ∑11+sin⁡2n\sum \frac{1}{1+\sin^2 n} converges.
  5. Converges because sin⁡2n\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+sin21+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 12

A model uses ∑n=2∞(−1)n1ln⁡n\sum_{n=2}^{\infty}(-1)^n\frac{1}{\ln n}. Does the series converge?

  1. Diverges because 1ln⁡n\frac{1}{\ln n} is increasing.
  2. Converges by the Alternating Series Test because 1ln⁡n\frac{1}{\ln n} decreases and approaches 00. (correct answer)
  3. Converges absolutely because ∑1ln⁡n\sum \frac{1}{\ln n} converges.
  4. Diverges because ∑1ln⁡n\sum \frac{1}{\ln n} diverges, so the alternating series diverges.
  5. Converges because lim⁡n→∞1ln⁡n=1\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 13

A series for a measurement error is ∑n=1∞(−1)n+1nn3+1\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{n^3+1}. Does it converge?

  1. Diverges because nn3+1\frac{n}{n^3+1} does not approach 00.
  2. Converges because ∑nn3+1\sum \frac{n}{n^3+1} diverges.
  3. Converges by the Alternating Series Test because nn3+1\frac{n}{n^3+1} decreases and approaches 00. (correct answer)
  4. Diverges because nn3+1\frac{n}{n^3+1} is increasing.
  5. Diverges because alternating series require bnb_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/(n3+1n^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 14

A series for corrections is ∑n=1∞(−1)n+11n(ln⁡n)2\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n(\ln n)^2} for n≥2n\ge2. Does it converge?

  1. Converges by the Alternating Series Test because 1n(ln⁡n)2\frac{1}{n(\ln n)^2} decreases and approaches 00. (correct answer)
  2. Diverges because 1n(ln⁡n)2\frac{1}{n(\ln n)^2} increases.
  3. Diverges because lim⁡n→∞1n(ln⁡n)2=1\lim_{n\to\infty}\frac{1}{n(\ln n)^2}=1.
  4. Converges because ∑1n(ln⁡n)2\sum \frac{1}{n(\ln n)^2} diverges.
  5. Diverges because alternating series require bnb_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} \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.

Question 15

A series for alternating work is ∑n=1∞(−1)narctan⁡(n)n\sum_{n=1}^{\infty}(-1)^n\frac{\arctan(n)}{n}. Does it converge?

  1. Converges by the Alternating Series Test because arctan⁡(n)n\frac{\arctan(n)}{n} decreases for large nn and approaches 00. (correct answer)
  2. Diverges because arctan⁡(n)\arctan(n) does not approach 00.
  3. Converges absolutely because ∑arctan⁡(n)n\sum \frac{\arctan(n)}{n} converges.
  4. Diverges because arctan⁡(n)n\frac{\arctan(n)}{n} is increasing.
  5. Diverges because alternating series require bnb_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 16

A Fourier-like term is ∑n=1∞(−1)n2n+1n\sum_{n=1}^{\infty}(-1)^{n}\frac{2n+1}{n}. Does it converge?

  1. Converges by the Alternating Series Test because 2n+1n\frac{2n+1}{n} decreases to 00.
  2. Diverges because lim⁡n→∞2n+1n≠0\lim_{n\to\infty}\frac{2n+1}{n}\neq 0. (correct answer)
  3. Converges because the terms alternate in sign.
  4. Converges absolutely because ∑2n+1n\sum \frac{2n+1}{n} converges.
  5. Diverges because 2n+1n\frac{2n+1}{n} is decreasing.
Explanation: The skill being tested is the Alternating Series Test for convergence. The test requires bn>0b_n > 0, decreasing, and approaching 0. Here, bn=2n+1n=2+1nb_n = \frac{2n+1}{n} = 2 + \frac{1}{n} approaches 2≠02 \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 bnb_n before assessing monotonicity in alternating series analysis.

Question 17

A power-series remainder uses ∑n=1∞(−1)n1n3\sum_{n=1}^{\infty}(-1)^n\frac{1}{n^3}. Does the series converge?

  1. Converges by the Alternating Series Test because 1n3\frac{1}{n^3} decreases and approaches 00. (correct answer)
  2. Diverges because ∑1n3\sum \frac{1}{n^3} diverges.
  3. Diverges because lim⁡n→∞1n3≠0\lim_{n\to\infty}\frac{1}{n^3}\neq 0.
  4. Converges because 1n3\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 18

A numerical approximation uses ∑n=1∞(−1)n+1n2n\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{2^n}. Does it converge?

  1. Converges by the Alternating Series Test because n2n\frac{n}{2^n} decreases for large nn and approaches 00. (correct answer)
  2. Diverges because n2n\frac{n}{2^n} increases without bound.
  3. Diverges because lim⁡n→∞n2n=1\lim_{n\to\infty}\frac{n}{2^n}=1.
  4. Converges because the terms do not alternate.
  5. Diverges because ∑n2n\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 19

A compensation sequence is ∑n=1∞(−1)n−11n\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{\sqrt{n}}. Does the series converge?

  1. Diverges because ∑1n\sum \frac{1}{\sqrt{n}} diverges, so the alternating series diverges.
  2. Converges because 1n\frac{1}{\sqrt{n}} decreases and approaches 00. (correct answer)
  3. Diverges because 1n\frac{1}{\sqrt{n}} is not decreasing for all nn.
  4. Converges because the ratio test limit equals 1/n1/\sqrt{n}.
  5. Diverges because alternating signs prevent convergence.
Explanation: This problem tests the alternating series test for convergence on ∑n=1∞(−1)n−11n\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{\sqrt{n}}. For the alternating series test, we need an=1na_n = \frac{1}{\sqrt{n}} to be eventually decreasing and approach zero. Since ddn(1n)=−12n3/2<0\frac{d}{dn}\left(\frac{1}{\sqrt{n}}\right) = -\frac{1}{2n^{3/2}} < 0 for all n>0n > 0, the function is decreasing, and clearly lim⁡n→∞1n=0\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 20

A signed error model is ∑n=1∞(−1)n1n+5\sum_{n=1}^{\infty}(-1)^n\frac{1}{n+5}. Does the series converge?

  1. Converges because 1n+5\frac{1}{n+5} decreases and approaches 00. (correct answer)
  2. Diverges because shifting by 55 changes convergence.
  3. Diverges because 1n+5\frac{1}{n+5} is not decreasing.
  4. Converges because the series is geometric with ratio −1-1.
  5. Diverges because ∑1n+5\sum \frac{1}{n+5} diverges, so the alternating one diverges.
Explanation: This question examines the alternating series test for ∑n=1∞(−1)n1n+5\sum_{n=1}^{\infty}(-1)^n\frac{1}{n+5}. The terms an=1n+5a_n = \frac{1}{n+5} clearly decrease since 1n+6<1n+5\frac{1}{n+6} < \frac{1}{n+5}, and lim⁡n→∞1n+5=0\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 ∑1n+5\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.