Calculus 2 Quiz: Alternating Series Test
20 questions · exam conditions
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Alternating Series TestQuestion 1 of 20

Which of the following alternating series diverges?

∑n=1∞(−1)nn+12n+1\sum_{n=1}^{\infty} (-1)^n \frac{n+1}{2n+1}
∑n=1∞(−1)n+1ln⁡(n+1)\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\ln(n+1)}
∑n=1∞(−1)ne−n\sum_{n=1}^{\infty} (-1)^n e^{-n}
∑n=1∞sin⁡((n+1/2)π)n\sum_{n=1}^{\infty} \frac{\sin((n+1/2)\pi)}{n}
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Calculus 2 Quiz

Calculus 2 Quiz: Alternating Series Test

Practice Alternating Series Test in Calculus 2 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, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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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

Which of the following alternating series diverges?

  1. ∑n=1∞(−1)nn+12n+1\sum_{n=1}^{\infty} (-1)^n \frac{n+1}{2n+1} (correct answer)
  2. ∑n=1∞(−1)n+1ln⁡(n+1)\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\ln(n+1)}
  3. ∑n=1∞(−1)ne−n\sum_{n=1}^{\infty} (-1)^n e^{-n}
  4. ∑n=1∞sin⁡((n+1/2)π)n\sum_{n=1}^{\infty} \frac{\sin((n+1/2)\pi)}{n}
Explanation: When you encounter questions about alternating series convergence, your primary tool is the Alternating Series Test (Leibniz criterion). This test states that an alternating series ∑(−1)nan\sum (-1)^n a_n converges if: (1) the terms ana_n are decreasing, and (2) lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0. However, if the limit condition fails, the series diverges by the Divergence Test. Let's examine each series by checking whether the terms approach zero: For choice A: ∑n=1∞(−1)nn+12n+1\sum_{n=1}^{\infty} (-1)^n \frac{n+1}{2n+1}, we have lim⁡n→∞n+12n+1=lim⁡n→∞1+1/n2+1/n=12≠0\lim_{n \to \infty} \frac{n+1}{2n+1} = \lim_{n \to \infty} \frac{1 + 1/n}{2 + 1/n} = \frac{1}{2} \neq 0. Since the terms don't approach zero, this series diverges by the Divergence Test. Choice B: ∑n=1∞(−1)n+1ln⁡(n+1)\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\ln(n+1)} has terms 1ln⁡(n+1)→0\frac{1}{\ln(n+1)} \to 0 as n→∞n \to \infty, and these terms are decreasing for large nn. This converges by the Alternating Series Test. Choice C: ∑n=1∞(−1)ne−n\sum_{n=1}^{\infty} (-1)^n e^{-n} has terms e−n→0e^{-n} \to 0 rapidly, and they're clearly decreasing. This converges. Choice D: ∑n=1∞sin⁡((n+1/2)π)n\sum_{n=1}^{\infty} \frac{\sin((n+1/2)\pi)}{n} can be rewritten as ∑n=1∞(−1)nn\sum_{n=1}^{\infty} \frac{(-1)^n}{n}, which is the alternating harmonic series. Since 1n→0\frac{1}{n} \to 0 and the terms decrease, this converges. Therefore, A is the only divergent series. Study tip: Always check the limit of terms first—if they don't approach zero, the series diverges immediately, regardless of the alternating pattern.

Question 2

For the series ∑n=2∞(−1)nnn\sum_{n=2}^{\infty} \frac{(-1)^n}{\sqrt[n]{n}}, what is the most complete analysis of convergence?

  1. Converges by the Alternating Series Test since 1nn→1\frac{1}{\sqrt[n]{n}} \to 1 and the sequence is eventually decreasing
  2. Diverges because lim⁡n→∞1nn=1≠0\lim_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1 \neq 0, violating the necessary condition for convergence (correct answer)
  3. Converges conditionally by the Alternating Series Test since 1nn→1\frac{1}{\sqrt[n]{n}} \to 1 but decreases to 1, not 0
  4. The Alternating Series Test is inconclusive; other tests must be used to determine convergence
Explanation: Since nn=n1/n\sqrt[n]{n} = n^{1/n} and lim⁡n→∞n1/n=1\lim_{n \to \infty} n^{1/n} = 1 (this can be shown using ln⁡(n1/n)=ln⁡nn→0\ln(n^{1/n}) = \frac{\ln n}{n} \to 0), we have lim⁡n→∞1nn=1≠0\lim_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1 \neq 0. By the divergence test (nth term test), the series ∑n=2∞(−1)nnn\sum_{n=2}^{\infty} \frac{(-1)^n}{\sqrt[n]{n}} diverges because its terms don't approach zero. The Alternating Series Test cannot even be applied since the first condition (terms approaching 0) fails. Choice A incorrectly states the limit is 0. Choice C makes the same error. Choice D suggests the AST is inconclusive when it actually cannot be applied at all.

Question 3

The series ∑n=1∞an\sum_{n=1}^{\infty} a_n is defined by an=(−1)nna_n = \frac{(-1)^n}{\sqrt{n}} if nn is odd, and an=(−1)nn2a_n = \frac{(-1)^n}{n^2} if nn is even. What is the behavior of this series?

  1. It diverges because it is the sum of a divergent series and a convergent series. (correct answer)
  2. It diverges because the sequence of magnitudes bn=∣an∣b_n = |a_n| is not monotonically decreasing.
  3. It converges by the Alternating Series Test, since bnb_n is eventually decreasing and its limit is 0.
  4. It converges absolutely because the terms approach zero faster than the harmonic series.
Explanation: When analyzing series with different patterns for odd and even terms, the key insight is to separate the series into its constituent parts and analyze each component's convergence behavior. This series can be written as the sum of two subseries: one containing all odd-indexed terms and one containing all even-indexed terms. For odd nn, we have an=(−1)nna_n = \frac{(-1)^n}{\sqrt{n}}, which gives us terms like −1,13,−15,…-1, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{5}}, \ldots. For even nn, we have an=(−1)nn2a_n = \frac{(-1)^n}{n^2}, which gives us terms like 14,116,136,…\frac{1}{4}, \frac{1}{16}, \frac{1}{36}, \ldots. The odd-indexed subseries behaves like −∑1odd numbers-\sum \frac{1}{\sqrt{\text{odd numbers}}}, which diverges since ∑1n\sum \frac{1}{\sqrt{n}} diverges (p-series with p=12<1p = \frac{1}{2} < 1). The even-indexed subseries converges absolutely since it's dominated by ∑1n2\sum \frac{1}{n^2}, which converges. Answer A correctly identifies that adding a divergent series to a convergent series yields divergence. Answer B incorrectly focuses on monotonicity of magnitudes, but this alone doesn't determine convergence of the entire series. Answer C misapplies the Alternating Series Test, which requires looking at the sequence {∣an∣}\{|a_n|\} as a whole, not separate patterns. Answer D incorrectly claims absolute convergence when the odd terms prevent this. Study tip: When series have different formulas for different indices, always separate them into subseries first, then analyze each part's convergence before combining the results.

Question 4

If the Alternating Series Test is used to establish the convergence of ∑n=1∞(−1)n+1n\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}, what is the minimum number of terms required to approximate the sum with an error less than 0.01?

  1. 9
  2. 10
  3. 99 (correct answer)
  4. 100
Explanation: The Alternating Series Estimation Theorem states that the error of the Nth partial sum, ∣S−SN∣|S - S_N|, is less than or equal to the magnitude of the first neglected term, bN+1b_{N+1}. We want ∣S−SN∣<0.01|S - S_N| < 0.01. We need to find NN such that bN+1<0.01b_{N+1} < 0.01. Here, bn=1/nb_n = 1/n, so we need bN+1=1N+1<0.01=1100b_{N+1} = \frac{1}{N+1} < 0.01 = \frac{1}{100}. This inequality becomes N+1>100N+1 > 100, which means N>99N > 99. The smallest integer value for NN is 100, but the question asks for the minimum number of terms in the sum, which is NN. To ensure the error is less than bN+1b_{N+1}, we need to sum up to the NNth term. If we sum N=99N=99 terms, the error is less than b100=1/100=0.01b_{100} = 1/100 = 0.01. So, 99 terms are required.

Question 5

If SN=∑n=1N(−1)n−1n2S_N = \sum_{n=1}^{N} \frac{(-1)^{n-1}}{n^2} is used to approximate S=∑n=1∞(−1)n−1n2S = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2}, what is the relationship between the true sum SS and the partial sum SNS_N?

  1. S>SNS > S_N for all even NN, and S<SNS < S_N for all odd NN. (correct answer)
  2. S<SNS < S_N for all even NN, and S>SNS > S_N for all odd NN.
  3. S>SNS > S_N for all NN.
  4. The relationship cannot be determined without computing more terms.
Explanation: For an alternating series where bnb_n is decreasing, the partial sums oscillate around the true sum SS. The first term is a1=1/12=1>0a_1 = 1/1^2 = 1 > 0. The odd partial sums S1,S3,...S_1, S_3, ... form a decreasing sequence that converges to SS from above, so SN>SS_N > S for odd NN. The even partial sums S2,S4,...S_2, S_4, ... form an increasing sequence that converges to SS from below, so SN<SS_N < S for even NN. Rearranging these inequalities gives S<SNS < S_N for odd NN and S>SNS > S_N for even NN.

Question 6

Let bnb_n be a positive, decreasing sequence with limit 0. The sum of the series S=∑n=1∞(−1)n+1bnS = \sum_{n=1}^{\infty} (-1)^{n+1}b_n is approximated by the second partial sum, S2=b1−b2S_2 = b_1 - b_2. Which of the following must be true about the true sum SS?

  1. S>b1S > b_1
  2. S<0S < 0
  3. S=b1−b2S = b_1 - b_2 exactly.
  4. b1−b2<S<b1b_1 - b_2 < S < b_1 (correct answer)
Explanation: The series converges by the AST. The sequence of partial sums oscillates around the true sum SS. The first term is b1b_1, so the first partial sum is S1=b1S_1 = b_1. The second term is −b2-b_2, so S2=b1−b2S_2 = b_1 - b_2. Since the terms are decreasing, b1−b2>0b_1 - b_2 > 0. The odd partial sums are overestimates and the even partial sums are underestimates. Therefore, S2<S<S1S_2 < S < S_1. Substituting the values for the partial sums gives b1−b2<S<b1b_1 - b_2 < S < b_1.

Question 7

Consider the Maclaurin series for ln⁡(1+x)\ln(1+x), which is ∑n=1∞(−1)n−1xnn\sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n}. For which value of xx in its interval of convergence does the series converge conditionally?

  1. x=−1x = -1
  2. x=1/2x = 1/2
  3. x=1x = 1 (correct answer)
  4. The series does not converge conditionally for any xx in its interval of convergence.
Explanation: The interval of convergence for the Maclaurin series for ln⁡(1+x)\ln(1+x) is (−1,1](-1, 1]. For xx in (−1,1)(-1, 1), the series converges absolutely. We must check the endpoints. At x=−1x = -1, the series is ∑(−1)n−1(−1)nn=∑−1n\sum (-1)^{n-1} \frac{(-1)^n}{n} = \sum \frac{-1}{n}, which is the negative of the harmonic series and diverges. At x=1x=1, the series becomes ∑(−1)n−11n\sum (-1)^{n-1} \frac{1}{n}, the alternating harmonic series. This series converges by the AST, but it does not converge absolutely since ∑1/n\sum 1/n diverges. Therefore, the series converges conditionally at x=1x=1.

Question 8

The series ∑n=1∞(−1)n+1n4\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^4} converges to a sum SS. If SS is approximated by the 4th partial sum, S4S_4, the Alternating Series Estimation Theorem states that the absolute error ∣S−S4∣|S - S_4| is bounded by what value?

  1. The magnitude of the 4th term, 1/2561/256.
  2. The magnitude of the 5th term, 1/6251/625. (correct answer)
  3. The value of the partial sum, S4S_4.
  4. A value that cannot be determined without calculating the exact sum SS.
Explanation: The Alternating Series Estimation Theorem states that for a convergent alternating series satisfying its conditions, the absolute error in approximating the total sum SS by the Nth partial sum SNS_N is less than or equal to the magnitude of the first neglected term, bN+1b_{N+1}. In this case, N=4N=4 and bn=1/n4b_n = 1/n^4. The error bound is ∣S−S4∣≤b4+1=b5=1/54=1/625|S - S_4| \le b_{4+1} = b_5 = 1/5^4 = 1/625.

Question 9

The series ∑n=1∞(−1)nn1/3n+1\sum_{n=1}^{\infty} (-1)^n \frac{n^{1/3}}{n+1} is being analyzed. What is the correct assessment?

  1. Converges by the Alternating Series Test since n1/3n+1→0\frac{n^{1/3}}{n+1} \to 0 and is eventually decreasing (correct answer)
  2. Diverges because lim⁡n→∞n1/3n+1=lim⁡n→∞1n2/3=0\lim_{n \to \infty} \frac{n^{1/3}}{n+1} = \lim_{n \to \infty} \frac{1}{n^{2/3}} = 0 but the sequence increases initially
  3. Converges absolutely since ∑n=1∞n1/3n+1\sum_{n=1}^{\infty} \frac{n^{1/3}}{n+1} converges by comparison with ∑n=1∞1n2/3\sum_{n=1}^{\infty} \frac{1}{n^{2/3}}
  4. Diverges because ∑n=1∞n1/3n+1\sum_{n=1}^{\infty} \frac{n^{1/3}}{n+1} diverges, and the alternating version cannot converge conditionally due to the growth rate
Explanation: We have lim⁡n→∞n1/3n+1=lim⁡n→∞n1/3n(1+1/n)=lim⁡n→∞1n2/3(1+1/n)=0\lim_{n \to \infty} \frac{n^{1/3}}{n+1} = \lim_{n \to \infty} \frac{n^{1/3}}{n(1+1/n)} = \lim_{n \to \infty} \frac{1}{n^{2/3}(1+1/n)} = 0. To check if the sequence is eventually decreasing, consider f(x)=x1/3x+1f(x) = \frac{x^{1/3}}{x+1}. Then f′(x)=13x−2/3(x+1)−x1/3(x+1)2=x−2/3(x+13−x)(x+1)2=x−2/3(1−2x3)(x+1)2f'(x) = \frac{\frac{1}{3}x^{-2/3}(x+1) - x^{1/3}}{(x+1)^2} = \frac{x^{-2/3}(\frac{x+1}{3} - x)}{(x+1)^2} = \frac{x^{-2/3}(\frac{1-2x}{3})}{(x+1)^2}. For large x, this is negative, so f is eventually decreasing. The Alternating Series Test confirms convergence. Choice B has the right limit but wrong conclusion. Choice C incorrectly claims absolute convergence. Choice D is wrong because conditional convergence is possible even when the absolute series diverges.

Question 10

The series ∑n=1∞an\sum_{n=1}^{\infty} a_n converges conditionally. Which of the following statements about the series ∑n=1∞(an+∣an∣)\sum_{n=1}^{\infty} (a_n + |a_n|) must be true?

  1. It converges absolutely.
  2. It converges conditionally.
  3. It diverges. (correct answer)
  4. Its convergence cannot be determined from the information given.
Explanation: If ∑an\sum a_n converges conditionally, then ∑an\sum a_n converges and ∑∣an∣\sum |a_n| diverges. Consider the new series ∑(an+∣an∣)\sum (a_n + |a_n|). If this series were to converge, then since ∑an\sum a_n converges, their difference must also converge. The difference is ∑(an+∣an∣)−∑an=∑∣an∣\sum (a_n + |a_n|) - \sum a_n = \sum |a_n|. This would imply that ∑∣an∣\sum |a_n| converges, which contradicts the definition of conditional convergence. Therefore, the series ∑(an+∣an∣)\sum (a_n + |a_n|) must diverge.

Question 11

Consider the series ∑n=2∞(−1)nn+(−1)nn2\sum_{n=2}^{\infty} (-1)^n \frac{n+(-1)^n}{n^2}. Which of the following statements accurately describes this series?

  1. The series converges by the Alternating Series Test because the terms are alternating and their limit is zero.
  2. The series diverges because the sequence of absolute values of the terms, bn=n+(−1)nn2b_n = \frac{n+(-1)^n}{n^2}, is not monotonically decreasing.
  3. The Alternating Series Test is inconclusive, but the series can be shown to converge by splitting it into two known convergent series. (correct answer)
  4. The Alternating Series Test is inconclusive, and the series can be shown to diverge by comparison with ∑n=2∞(−1)nn\sum_{n=2}^{\infty} \frac{(-1)^n}{n}.
Explanation: The series is alternating. Let bn=n+(−1)nn2b_n = \frac{n+(-1)^n}{n^2}. The sequence {bn}\{b_n\} is not monotonically decreasing (e.g., b3=2/9≈0.222b_3 = 2/9 \approx 0.222 and b4=5/16=0.3125b_4 = 5/16 = 0.3125). Therefore, the Alternating Series Test is inconclusive. However, the series can be rewritten as ∑n=2∞((−1)nnn2+(−1)n(−1)nn2)=∑n=2∞(−1)nn+∑n=2∞1n2\sum_{n=2}^{\infty} \left( \frac{(-1)^n n}{n^2} + \frac{(-1)^n (-1)^n}{n^2} \right) = \sum_{n=2}^{\infty} \frac{(-1)^n}{n} + \sum_{n=2}^{\infty} \frac{1}{n^2}. The first series is a convergent alternating series, and the second is a convergent p-series. The sum of two convergent series is convergent.

Question 12

Which of the following statements is true for the series ∑n=1∞cos⁡(nπ)n1/2+1\sum_{n=1}^{\infty} \frac{\cos(n\pi)}{n^{1/2} + 1}?

  1. The series diverges because cos⁡(nπ)\cos(n\pi) oscillates.
  2. The series converges absolutely.
  3. The series converges conditionally. (correct answer)
  4. The Alternating Series Test is inconclusive for this series.
Explanation: The term cos⁡(nπ)\cos(n\pi) is equivalent to (−1)n(-1)^n. So the series is ∑n=1∞(−1)nn+1\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n} + 1}. This is an alternating series with bn=1n+1b_n = \frac{1}{\sqrt{n} + 1}. Since lim⁡n→∞bn=0\lim_{n\to\infty} b_n = 0 and bnb_n is a decreasing sequence, the series converges by the Alternating Series Test. To check for absolute convergence, we examine ∑n=1∞1n+1\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} + 1}. By the Limit Comparison Test with the divergent p-series ∑1n\sum \frac{1}{\sqrt{n}} (where p=1/2p=1/2), this series diverges. Since the original series converges but not absolutely, it converges conditionally.

Question 13

Let ∑n=1∞(−1)n+1bn\sum_{n=1}^\infty (-1)^{n+1} b_n be a series where bn>0b_n > 0. Which of the following conditions, by itself, is sufficient to guarantee the series converges?

  1. bn+1≤bnb_{n+1} \le b_n for all n≥1n \ge 1.
  2. lim⁡n→∞bn=0\lim_{n\to\infty} b_n = 0.
  3. The series ∑n=1∞bn\sum_{n=1}^\infty b_n converges. (correct answer)
  4. bn=f(n)b_n = f(n) where f(x)f(x) is a positive, differentiable function with f′(x)<0f'(x) < 0 for x≥1x \ge 1.
Explanation: If the series of absolute values, ∑bn\sum b_n, converges, then the series is absolutely convergent. A fundamental theorem of series states that if a series converges absolutely, then it converges. Therefore, condition C is sufficient. Conditions A and D only establish that the terms are decreasing, which is not sufficient without knowing the limit. Condition B establishes that the limit of the terms is zero, which is not sufficient without the terms being eventually decreasing.

Question 14

Which statement correctly describes the series ∑n=2∞(−1)nln⁡(n2)n\sum_{n=2}^{\infty} (-1)^n \frac{\ln(n^2)}{n}?

  1. The series converges absolutely.
  2. The series converges conditionally. (correct answer)
  3. The series diverges because ln⁡(n2)\ln(n^2) grows without bound.
  4. The series diverges because its terms are not monotonically decreasing.
Explanation: The series can be written as ∑n=2∞(−1)n2ln⁡nn\sum_{n=2}^{\infty} (-1)^n \frac{2\ln n}{n}. Let bn=2ln⁡nnb_n = \frac{2\ln n}{n}. Using L'Hôpital's Rule, lim⁡n→∞bn=lim⁡n→∞2/n1=0\lim_{n\to\infty} b_n = \lim_{n\to\infty} \frac{2/n}{1} = 0. To check if bnb_n is decreasing, consider the function f(x)=2ln⁡xxf(x) = \frac{2\ln x}{x}. Its derivative is f′(x)=2(1−ln⁡x)x2f'(x) = \frac{2(1-\ln x)}{x^2}, which is negative for x>ex > e. Thus, the terms are eventually decreasing. By the AST, the series converges. For absolute convergence, we examine ∑2ln⁡nn\sum \frac{2\ln n}{n}. Since ln⁡n>1\ln n > 1 for n>en > e, we have 2ln⁡nn>2n\frac{2\ln n}{n} > \frac{2}{n}. Since ∑2n\sum \frac{2}{n} diverges, ∑2ln⁡nn\sum \frac{2\ln n}{n} diverges by comparison. Therefore, the series converges conditionally.

Question 15

Let f(x)f(x) be a positive, continuous, and decreasing function for x≥1x \ge 1 such that ∫1∞f(x)dx\int_1^\infty f(x) dx diverges. What can be concluded about the series ∑n=1∞(−1)nf(n)\sum_{n=1}^\infty (-1)^n f(n)?

  1. It must diverge.
  2. It must converge absolutely.
  3. It converges, but it might be conditional or absolute.
  4. It must converge conditionally, provided lim⁡x→∞f(x)=0\lim_{x\to\infty} f(x) = 0. (correct answer)
Explanation: Let bn=f(n)b_n = f(n). The conditions given are that bnb_n is positive and decreasing. For the Alternating Series Test to apply, we also need lim⁡n→∞bn=0\lim_{n\to\infty} b_n = 0. The problem implicitly suggests this by framing it as a standard convergence problem. If we assume lim⁡n→∞f(n)=0\lim_{n\to\infty} f(n) = 0, the series ∑(−1)nf(n)\sum (-1)^n f(n) converges by the AST. The Integral Test states that since ∫1∞f(x)dx\int_1^\infty f(x) dx diverges, the series of absolute values, ∑f(n)\sum f(n), must also diverge. A series that converges but does not converge absolutely is conditionally convergent. The condition lim⁡x→∞f(x)=0\lim_{x\to\infty} f(x) = 0 is necessary for this conclusion.

Question 16

The Alternating Series Test can be used to show convergence for which of the following series?

  1. ∑n=1∞(−1)n+1n\sum_{n=1}^{\infty} \frac{(-1)^n + 1}{n}
  2. ∑n=1∞(−1)nnn+1\sum_{n=1}^{\infty} (-1)^n \frac{n}{n+1}
  3. ∑n=1∞(−1)n2+sin⁡nn2\sum_{n=1}^{\infty} (-1)^n \frac{2+\sin n}{n^2}
  4. ∑n=1∞(−1)nnen\sum_{n=1}^{\infty} (-1)^n \frac{n}{e^n} (correct answer)
Explanation: A) This series is 0+2/2+0+2/4+...0 + 2/2 + 0 + 2/4 + ... which is the harmonic series, not an alternating series. AST does not apply. B) The limit of the terms bn=n/(n+1)b_n = n/(n+1) is 1, not 0. Series diverges by nth-Term Test. AST does not show convergence. C) The terms bn=(2+sin⁡n)/n2b_n = (2+\sin n)/n^2 are not guaranteed to be decreasing due to the sin⁡n\sin n term. AST cannot be directly applied. D) Let bn=n/enb_n = n/e^n. lim⁡n→∞n/en=0\lim_{n\to\infty} n/e^n = 0 by L'Hôpital's Rule. The function f(x)=x/exf(x) = x/e^x has derivative f′(x)=(1−x)/exf'(x) = (1-x)/e^x, which is negative for x>1x>1. So, the terms are decreasing for n≥2n\ge 2. Both conditions of the AST are met.

Question 17

What is the behavior of the series ∑n=1∞sin⁡((2n−1)π2)1n\sum_{n=1}^{\infty} \sin\left(\frac{(2n-1)\pi}{2}\right) \frac{1}{\sqrt{n}}?

  1. The series converges conditionally. (correct answer)
  2. The series is not an alternating series, so the Alternating Series Test cannot be applied.
  3. The series diverges because sin⁡(x)\sin(x) oscillates.
  4. The series converges absolutely.
Explanation: When you encounter a series with trigonometric terms, you need to analyze both the trigonometric behavior and apply appropriate convergence tests. This series combines a trigonometric function with a power term, making it a candidate for alternating series analysis. First, let's examine sin⁡((2n−1)π2)\sin\left(\frac{(2n-1)\pi}{2}\right). For odd values of 2n−12n-1, this gives us sin⁡(π/2)=1\sin(\pi/2) = 1, sin⁡(3π/2)=−1\sin(3\pi/2) = -1, sin⁡(5π/2)=1\sin(5\pi/2) = 1, and so on. The pattern alternates: 1,−1,1,−1,...1, -1, 1, -1, ..., so our series becomes ∑n=1∞(−1)n+11n\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{\sqrt{n}} This is indeed an alternating series where the terms 1n\frac{1}{\sqrt{n}} decrease monotonically to zero. By the Alternating Series Test, the series converges. However, to check absolute convergence, we examine ∑n=1∞∣sin⁡((2n−1)π2)∣1n=∑n=1∞1n\sum_{n=1}^{\infty} \left|\sin\left(\frac{(2n-1)\pi}{2}\right)\right| \frac{1}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} This is a p-series with p=1/2<1p = 1/2 < 1, which diverges. Since the original series converges but its absolute value diverges, the series converges conditionally. Option B is wrong because this clearly is an alternating series. Option C incorrectly suggests that oscillation alone causes divergence—oscillation can actually help convergence in alternating series. Option D is wrong because absolute convergence requires the p-series to converge, which it doesn't. Strategy tip: For series with trigonometric terms, always evaluate the trig function at specific values to identify patterns, then check both conditional and absolute convergence separately.

Question 18

Let ∑n=1∞(−1)nbn\sum_{n=1}^{\infty} (-1)^n b_n be a series with bn>0b_n > 0. If lim⁡n→∞bn=L\lim_{n\to\infty} b_n = L where LL is a finite positive number, what can be definitively concluded about the series?

  1. The series converges conditionally because the terms alternate and approach a finite limit.
  2. The Alternating Series Test is inconclusive because the terms are not guaranteed to be decreasing.
  3. The series converges absolutely by comparison with a constant series.
  4. The series diverges by the nth-Term Test for Divergence. (correct answer)
Explanation: If lim⁡n→∞bn=L>0\lim_{n\to\infty} b_n = L > 0, then the limit of the terms of the series, lim⁡n→∞(−1)nbn\lim_{n\to\infty} (-1)^n b_n, does not exist because the terms oscillate between values close to LL and −L-L. Since the limit of the terms is not 0, the series diverges by the nth-Term Test for Divergence. The Alternating Series Test is not merely inconclusive; a definitive conclusion of divergence can be made.

Question 19

Consider the series ∑n=1∞(−1)n(n+1−n)\sum_{n=1}^{\infty} (-1)^n (\sqrt{n+1} - \sqrt{n}). What is the behavior of this series?

  1. Diverges because n+1\sqrt{n+1} and n\sqrt{n} both diverge to infinity.
  2. Converges conditionally. (correct answer)
  3. Converges absolutely.
  4. Diverges by the nth-Term Test.
Explanation: Let bn=n+1−nb_n = \sqrt{n+1} - \sqrt{n}. We can rationalize this expression: bn=(n+1−n)n+1+nn+1+n=n+1−nn+1+n=1n+1+nb_n = (\sqrt{n+1} - \sqrt{n}) \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}} = \frac{n+1-n}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}. The sequence {bn}\{b_n\} is positive, decreasing, and lim⁡n→∞bn=0\lim_{n\to\infty} b_n = 0. By the AST, the series converges. To test for absolute convergence, we examine ∑bn=∑1n+1+n\sum b_n = \sum \frac{1}{\sqrt{n+1} + \sqrt{n}}. We can use the Limit Comparison Test with the divergent p-series ∑1n\sum \frac{1}{\sqrt{n}}. The limit of the ratio of terms is 1/21/2. Since the p-series diverges, ∑bn\sum b_n also diverges. Thus, the series converges conditionally.

Question 20

The series ∑n=1∞(−10)nn!\sum_{n=1}^{\infty} \frac{(-10)^n}{n!} is an alternating series. Which of the following is true?

  1. The series diverges because the terms (−10)n(-10)^n grow in magnitude.
  2. The series converges conditionally by the Alternating Series Test.
  3. The series converges absolutely by the Ratio Test. (correct answer)
  4. The Alternating Series Test is inconclusive because 10nn!\frac{10^n}{n!} is not decreasing for small nn.
Explanation: To determine the behavior of the series, we can test for absolute convergence using the Ratio Test on ∑n=1∞10nn!\sum_{n=1}^{\infty} \frac{10^n}{n!}. The ratio of successive terms is ∣an+1an∣=10n+1(n+1)!⋅n!10n=10n+1|\frac{a_{n+1}}{a_n}| = \frac{10^{n+1}}{(n+1)!} \cdot \frac{n!}{10^n} = \frac{10}{n+1}. Taking the limit as n→∞n \to \infty, we get lim⁡n→∞10n+1=0\lim_{n\to\infty} \frac{10}{n+1} = 0. Since the limit is less than 1, the series converges absolutely. Because absolute convergence implies convergence, the original series converges.