Calculus 2 Quiz: Convergent And Divergent Series
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Convergent And Divergent SeriesQuestion 1 of 20

Suppose ∑n=1∞∣an∣\sum_{n=1}^{\infty} |a_n| converges. Which of the following MUST be true about the series ∑n=1∞an\sum_{n=1}^{\infty} a_n?

The series ∑an\sum a_n diverges.
The series ∑an\sum a_n converges to 0.
The series ∑an\sum a_n converges.
The series ∑an\sum a_n could either converge or diverge.
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Calculus 2 Quiz

Calculus 2 Quiz: Convergent And Divergent Series

Practice Convergent And Divergent Series 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 Convergent And Divergent Series, 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

Suppose ∑n=1∞∣an∣\sum_{n=1}^{\infty} |a_n| converges. Which of the following MUST be true about the series ∑n=1∞an\sum_{n=1}^{\infty} a_n?

  1. The series ∑an\sum a_n diverges.
  2. The series ∑an\sum a_n converges to 0.
  3. The series ∑an\sum a_n converges. (correct answer)
  4. The series ∑an\sum a_n could either converge or diverge.
Explanation: This question relates to the concept of absolute convergence. A fundamental theorem of infinite series states that if a series converges absolutely (i.e., ∑∣an∣\sum |a_n| converges), then the series itself (∑an\sum a_n) must also converge. The sum is not necessarily 0. This is because 0≤an+∣an∣≤2∣an∣0 \le a_n + |a_n| \le 2|a_n|. By comparison, ∑(an+∣an∣)\sum (a_n + |a_n|) converges. Since ∑an=∑(an+∣an∣)−∑∣an∣\sum a_n = \sum (a_n + |a_n|) - \sum |a_n|, it is the difference of two convergent series and must therefore converge.

Question 2

Let f(x)f(x) be a function such that f(1)=10f(1)=10 and lim⁡x→∞f(x)=2\lim_{x\to\infty} f(x) = 2. The series ∑n=1∞(f(n)−f(n+1))\sum_{n=1}^\infty (f(n) - f(n+1)) is a telescoping series. What is the sum of this series?

  1. 8 (correct answer)
  2. 10
  3. 12
  4. The series diverges.
Explanation: The Nth partial sum of this telescoping series is SN=(f(1)−f(2))+(f(2)−f(3))+⋯+(f(N)−f(N+1))S_N = (f(1) - f(2)) + (f(2) - f(3)) + \dots + (f(N) - f(N+1)). The intermediate terms cancel out, leaving SN=f(1)−f(N+1)S_N = f(1) - f(N+1). To find the sum of the infinite series, we take the limit of SNS_N as N→∞N \to \infty. Sum=lim⁡N→∞SN=lim⁡N→∞(f(1)−f(N+1))=f(1)−lim⁡N→∞f(N+1).\text{Sum} = \lim_{N \to \infty} S_N = \lim_{N \to \infty} (f(1) - f(N+1)) = f(1) - \lim_{N \to \infty} f(N+1). Given f(1)=10f(1)=10 and lim⁡x→∞f(x)=2\lim_{x\to\infty} f(x) = 2, the sum is 10−2=810 - 2 = 8.

Question 3

A series ∑n=1∞an\sum_{n=1}^{\infty} a_n converges to a sum SS. A new series is formed, ∑k=5∞ak−2\sum_{k=5}^{\infty} a_{k-2}. What is the sum of this new series in terms of SS, a1a_1, and a2a_2?

  1. S−a1−a2S - a_1 - a_2 (correct answer)
  2. S−a3−a4S - a_3 - a_4
  3. S−a1−a2−a3−a4S - a_1 - a_2 - a_3 - a_4
  4. The sum cannot be determined without knowing the terms.
Explanation: Let's re-index the new series. Let n=k−2n = k-2. When the starting index is k=5k=5, the new index is n=5−2=3n = 5-2 = 3. As k→∞k \to \infty, n→∞n \to \infty. So the new series is ∑n=3∞an\sum_{n=3}^{\infty} a_n. This is the original series with the first two terms removed. Therefore, its sum is the original sum SS minus the first two terms, a1a_1 and a2a_2. The sum is S−(a1+a2)S - (a_1 + a_2).

Question 4

Let ∑n=1∞an\sum_{n=1}^{\infty} a_n be a convergent series. Let RN=∑n=N+1∞anR_N = \sum_{n=N+1}^{\infty} a_n be the remainder of the series after the Nth term. Which of the following must be true?

  1. lim⁡N→∞RN=∑n=1∞an\lim_{N \to \infty} R_N = \sum_{n=1}^{\infty} a_n
  2. lim⁡N→∞RN=0\lim_{N \to \infty} R_N = 0 (correct answer)
  3. RNR_N must be a decreasing sequence.
  4. The sequence {RN}\{R_N\} diverges.
Explanation: Let the sum of the series be SS. By definition, S=lim⁡N→∞SNS = \lim_{N \to \infty} S_N, where SNS_N is the Nth partial sum. The sum SS can be written as S=SN+RNS = S_N + R_N. Therefore, RN=S−SNR_N = S - S_N. As N→∞N \to \infty, SNS_N approaches SS. So, lim⁡N→∞RN=lim⁡N→∞(S−SN)=S−S=0\lim_{N \to \infty} R_N = \lim_{N \to \infty} (S - S_N) = S - S = 0. The remainder must approach zero for a series to converge.

Question 5

Let S=∑n=1∞(−1)n+11n=1−12+13−14+…S = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots. It is known that S=ln⁡(2)S = \ln(2). Which statement about the sequence of partial sums {SN}\{S_N\} is correct?

  1. The sequence {SN}\{S_N\} is monotonically increasing towards ln⁡(2)\ln(2).
  2. The sequence {SN}\{S_N\} oscillates, with terms alternately above and below ln⁡(2)\ln(2).
  3. The sequence {SN}\{S_N\} is bounded by 0 and 1, and converges to ln⁡(2)\ln(2). (correct answer)
  4. The sequence {SN}\{S_N\} diverges, even though the series has a defined value.
Explanation: Let's examine the first few partial sums: S1=1S_1 = 1, S2=1−1/2=0.5S_2 = 1 - 1/2 = 0.5, S3=0.5+1/3≈0.833S_3 = 0.5 + 1/3 \approx 0.833, S4≈0.833−1/4=0.583S_4 \approx 0.833 - 1/4 = 0.583. The sequence of partial sums is not monotonic; it oscillates. However, the oscillations get smaller and hone in on the limit. The odd partial sums form a decreasing sequence (S1>S3>S5…S_1 > S_3 > S_5 \dots) and the even partial sums form an increasing sequence (S2<S4<S6…S_2 < S_4 < S_6 \dots). All terms are bounded between S2=0.5S_2=0.5 and S1=1S_1=1. Since ln⁡(2)≈0.693\ln(2) \approx 0.693, the sequence converges to this value within the bounds [0.5, 1]. Option C is the most accurate description.

Question 6

Which of the following statements about infinite series is FALSE?

  1. If ∑an\sum a_n converges, then lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0.
  2. If {SN}\{S_N\} is the sequence of partial sums for ∑an\sum a_n and lim⁡N→∞SN=2\lim_{N \to \infty} S_N = 2, then ∑an\sum a_n converges.
  3. If lim⁡n→∞an=L≠0\lim_{n \to \infty} a_n = L \neq 0, then ∑an\sum a_n diverges.
  4. If lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0, then ∑an\sum a_n converges. (correct answer)
Explanation: Statement D is the converse of the n-th Term Test for Divergence, and it is false. While lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0 is a necessary condition for convergence, it is not sufficient. The classic counterexample is the harmonic series ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}. For this series, lim⁡n→∞1n=0\lim_{n \to \infty} \frac{1}{n} = 0, but the series diverges.

Question 7

Suppose ∑n=1∞an\sum_{n=1}^{\infty} a_n is a series with positive terms and its sequence of partial sums is SNS_N. Which condition is sufficient to guarantee the convergence of the series?

  1. lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0.
  2. The sequence {an}\{a_n\} is decreasing.
  3. The sequence {SN}\{S_N\} is bounded above. (correct answer)
  4. The sequence {SN}\{S_N\} is decreasing.
Explanation: Since the series has positive terms (an>0a_n > 0), the sequence of partial sums SN=SN−1+aNS_N = S_{N-1} + a_N is strictly increasing. The Monotone Convergence Theorem states that a sequence that is both monotonic and bounded must converge. Since {SN}\{S_N\} is already monotonic (increasing), the only additional condition needed to guarantee convergence is that it must be bounded. Because it's increasing, it is already bounded below by S1S_1, so being bounded above is sufficient. A is necessary but not sufficient. B is not sufficient (e.g., harmonic series). D is impossible since an>0a_n > 0.

Question 8

Let ∑an\sum a_n and ∑bn\sum b_n be series with positive terms. If ∑an\sum a_n converges and ∑bn\sum b_n diverges, which of the following series must diverge?

  1. ∑(anbn)\sum (a_n b_n)
  2. ∑anbn\sum \frac{a_n}{b_n}
  3. ∑(an+bn)\sum (a_n + b_n) (correct answer)
  4. ∑max⁡(an,bn)\sum \max(a_n, b_n)
Explanation: If ∑an\sum a_n converges to LL and ∑bn\sum b_n diverges, then ∑(an+bn)\sum (a_n + b_n) must diverge. If it were to converge to a sum MM, then ∑bn=∑((an+bn)−an)=M−L\sum b_n = \sum ((a_n+b_n) - a_n) = M - L, which would imply ∑bn\sum b_n converges, contradicting our assumption. For counterexamples to other choices: If an=1/n2a_n = 1/n^2 and bn=1/nb_n = 1/n, then ∑anbn=∑1/n3\sum a_n b_n = \sum 1/n^3 converges. If an=1/n2a_n = 1/n^2 and bn=nb_n = n, then ∑an/bn=∑1/n3\sum a_n/b_n = \sum 1/n^3 converges. Choice D can also converge in some cases.

Question 9

Given that ∑n=1∞an\sum_{n=1}^{\infty} a_n converges and an>0a_n > 0 for all nn. Which of the following statements must be true?

  1. The series ∑n=1∞an\sum_{n=1}^{\infty} \sqrt{a_n} converges.
  2. The series ∑n=1∞1an\sum_{n=1}^{\infty} \frac{1}{a_n} diverges. (correct answer)
  3. The sequence of partial sums is decreasing and bounded.
  4. The sequence {an}\{a_n\} is strictly decreasing.
Explanation: If ∑an\sum a_n converges, then it is necessary that lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0. Since an>0a_n > 0, we can consider the terms of the series ∑1/an\sum 1/a_n. The limit of these terms is lim⁡n→∞1an\lim_{n \to \infty} \frac{1}{a_n}, which approaches +∞+\infty as ana_n approaches 0. By the n-th Term Test for Divergence, since the limit of the terms is not zero, the series ∑1/an\sum 1/a_n must diverge. For A, a counterexample is an=1/n2a_n = 1/n^2; ∑an\sum a_n converges but ∑an=∑1/n\sum \sqrt{a_n} = \sum 1/n diverges. C is false because with an>0a_n > 0, the partial sums are increasing. D is not necessarily true.

Question 10

Consider a series ∑n=1∞an\sum_{n=1}^{\infty} a_n where the sequence of partial sums is given by SN=arctan⁡(N)S_N = \arctan(N). Which of the following is true?

  1. The series converges to π/2\pi/2. (correct answer)
  2. The series converges to 0.
  3. The series converges to 1.
  4. The series diverges.
Explanation: The sum of a series is defined as the limit of its sequence of partial sums, if that limit exists and is finite. Here, the Nth partial sum is SN=arctan⁡(N)S_N = \arctan(N). We need to find the limit of SNS_N as N→∞N \to \infty. lim⁡N→∞SN=lim⁡N→∞arctan⁡(N)=π2.\lim_{N \to \infty} S_N = \lim_{N \to \infty} \arctan(N) = \frac{\pi}{2}. Since the limit is the finite value π/2\pi/2, the series converges to π/2\pi/2.

Question 11

Which of the following series can be shown to be divergent using only the n-th Term Test for Divergence?

  1. ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{n}
  2. ∑n=1∞ln⁡nn\sum_{n=1}^{\infty} \frac{\ln n}{n}
  3. ∑n=1∞nsin⁡(1/n)\sum_{n=1}^{\infty} n \sin(1/n) (correct answer)
  4. ∑n=1∞nn2+1\sum_{n=1}^{\infty} \frac{n}{n^2+1}
Explanation: The n-th Term Test states that if lim⁡n→∞an≠0\lim_{n \to \infty} a_n \neq 0, the series diverges. We test each option: A) lim⁡n→∞1/n=0\lim_{n \to \infty} 1/n = 0. Test is inconclusive. B) lim⁡n→∞(ln⁡n)/n=0\lim_{n \to \infty} (\ln n)/n = 0 by L'Hopital's Rule. Test is inconclusive. C) lim⁡n→∞nsin⁡(1/n)=lim⁡x→0+sin⁡xx=1\lim_{n \to \infty} n \sin(1/n) = \lim_{x \to 0^+} \frac{\sin x}{x} = 1 (by letting x=1/nx=1/n). Since the limit is 1≠01 \neq 0, the series diverges by the n-th Term Test. D) lim⁡n→∞n/(n2+1)=0\lim_{n \to \infty} n/(n^2+1) = 0. Test is inconclusive.

Question 12

Find the sum of the infinite series ∑n=1∞2n(n+2)\sum_{n=1}^{\infty} \frac{2}{n(n+2)}.

  1. 1
  2. 3/2 (correct answer)
  3. 2
  4. The series diverges.
Explanation: This is a telescoping series. First, use partial fraction decomposition: 2n(n+2)=An+Bn+2\frac{2}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2}. Solving gives A=1A=1 and B=−1B=-1, so the term is 1n−1n+2\frac{1}{n} - \frac{1}{n+2}. The Nth partial sum SNS_N is: SN=(1−13)+(12−14)+(13−15)+⋯+(1N−1N+2)S_N = \left(1 - \frac{1}{3}\right) + \left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \dots + \left(\frac{1}{N} - \frac{1}{N+2}\right) Most terms cancel, leaving SN=1+12−1N+1−1N+2S_N = 1 + \frac{1}{2} - \frac{1}{N+1} - \frac{1}{N+2}. The sum of the series is lim⁡N→∞SN=1+12−0−0=32\lim_{N \to \infty} S_N = 1 + \frac{1}{2} - 0 - 0 = \frac{3}{2}.

Question 13

Let ∑n=1∞an\sum_{n=1}^{\infty} a_n be an infinite series with its Nth partial sum given by SN=3N2−N2N2+5S_N = \frac{3N^2 - N}{2N^2 + 5}. What is the sum of the series?

  1. The sum is 0.
  2. The sum is 3/23/2. (correct answer)
  3. The sum is 5/25/2.
  4. The series diverges.
Explanation: By definition, the sum of a convergent infinite series is the limit of its sequence of partial sums. We need to compute lim⁡N→∞SN\lim_{N \to \infty} S_N. $$ \lim_{N \to \infty} S_N = \lim_{N \to \infty} \frac{3N^2 - N}{2N^2 + 5} = \lim_{N \to \infty} \frac{N^2(3 - 1/N)}{N^2(2 + 5/N^2)} = \frac{3 - 0}{2 + 0} = \frac{3}{2}.

Question 14

If the infinite series ∑n=1∞an\sum_{n=1}^{\infty} a_n converges to a finite sum LL, which statement regarding its sequence of partial sums, {SN}\{S_N\}, must be true?

  1. The sequence {SN}\{S_N\} is always monotonic and bounded.
  2. The sequence {SN}\{S_N\} must converge to 0.
  3. For any ϵ>0\epsilon > 0, there exists an integer MM such that for all N>MN > M, ∣SN−L∣<ϵ|S_N - L| < \epsilon. (correct answer)
  4. The difference between consecutive partial sums, SN−SN−1S_N - S_{N-1}, approaches LL as N→∞N \to \infty.
Explanation: The statement that the series ∑an\sum a_n converges to LL is, by definition, equivalent to the statement that its sequence of partial sums {SN}\{S_N\} converges to LL. Choice C is the formal definition of the limit lim⁡N→∞SN=L\lim_{N \to \infty} S_N = L. Choice A is only true if all terms ana_n have the same sign. Choice B confuses the limit of the partial sums with the limit of the terms ana_n. Choice D is incorrect; SN−SN−1=aNS_N - S_{N-1} = a_N, and lim⁡N→∞aN=0\lim_{N \to \infty} a_N = 0, not LL.

Question 15

Suppose ∑n=1∞an\sum_{n=1}^{\infty} a_n is a convergent series and ∑n=1∞bn\sum_{n=1}^{\infty} b_n is a divergent series. What can be concluded about the series ∑n=1∞(an−bn)\sum_{n=1}^{\infty} (a_n - b_n)?

  1. It must converge.
  2. It must diverge. (correct answer)
  3. It could either converge or diverge, depending on the specific series.
  4. It converges if and only if lim⁡n→∞bn=0\lim_{n\to\infty} b_n = 0.
Explanation: The difference of a convergent series and a divergent series must be divergent. This can be shown by contradiction. Assume ∑(an−bn)\sum (a_n - b_n) converges to a sum LL. Since ∑an\sum a_n converges to a sum SS, we could write ∑bn=∑(an−(an−bn))\sum b_n = \sum (a_n - (a_n - b_n)). By the properties of convergent series, this would be the difference of two convergent series, which must converge to S−LS - L. This contradicts the given information that ∑bn\sum b_n diverges. Therefore, ∑(an−bn)\sum (a_n - b_n) must diverge.

Question 16

Let SN=∑n=1NanS_N = \sum_{n=1}^{N} a_n. If the sequence of partial sums {SN}\{S_N\} is given by SN=5−(N+1)e−NS_N = 5 - (N+1)e^{-N}, what is a2a_2?

  1. 3e−23e^{-2}
  2. 5−3e−25 - 3e^{-2}
  3. e−1−3e−2e^{-1} - 3e^{-2}
  4. 2e−1−3e−22e^{-1} - 3e^{-2} (correct answer)
Explanation: The n-th term of the series can be found using the relation an=Sn−Sn−1a_n = S_n - S_{n-1} for n≥2n \ge 2. We need to find a2a_2, which is S2−S1S_2 - S_1. First, calculate S2S_2: S2=5−(2+1)e−2=5−3e−2S_2 = 5 - (2+1)e^{-2} = 5 - 3e^{-2}. Next, calculate S1S_1: S1=5−(1+1)e−1=5−2e−1S_1 = 5 - (1+1)e^{-1} = 5 - 2e^{-1}. Then, a2=S2−S1=(5−3e−2)−(5−2e−1)=2e−1−3e−2a_2 = S_2 - S_1 = (5 - 3e^{-2}) - (5 - 2e^{-1}) = 2e^{-1} - 3e^{-2}.

Question 17

The series ∑n=1∞c⋅an\sum_{n=1}^{\infty} c \cdot a_n converges if ∑n=1∞an\sum_{n=1}^{\infty} a_n converges. What must be true about the constant cc?

  1. cc can be any real number. (correct answer)
  2. cc must be a non-zero real number.
  3. cc must be a positive real number.
  4. cc must satisfy ∣c∣<1|c| < 1.
Explanation: One of the fundamental properties of convergent series is that if ∑an\sum a_n converges to a sum SS, then for any real constant cc, the series ∑c⋅an\sum c \cdot a_n also converges, and its sum is c⋅Sc \cdot S. This property holds for any real number cc, including c=0c=0 (in which case the series converges to 0). The condition is not restricted to non-zero, positive, or small values of cc.

Question 18

For what values of the real number xx does the infinite series ∑n=1∞(ln⁡x)n\sum_{n=1}^{\infty} (\ln x)^n converge?

  1. x>0x > 0
  2. ∣x∣<e|x| < e
  3. 1/e<x<e1/e < x < e (correct answer)
  4. 0<x<e0 < x < e
Explanation: This is a geometric series with common ratio r=ln⁡xr = \ln x. A geometric series converges if and only if ∣r∣<1|r| < 1. In this case, we require ∣ln⁡x∣<1|\ln x| < 1. This inequality is equivalent to −1<ln⁡x<1-1 < \ln x < 1. To solve for xx, we apply the exponential function to all parts of the inequality: e−1<eln⁡x<e1e^{-1} < e^{\ln x} < e^1, which simplifies to 1/e<x<e1/e < x < e.

Question 19

Let an=f(n)a_n = f(n) be a sequence of positive, decreasing terms. If the series ∑n=1∞an\sum_{n=1}^{\infty} a_n converges, what can be said about ∑n=1∞an+1\sum_{n=1}^{\infty} a_{n+1}?

  1. It converges to the same sum as ∑n=1∞an\sum_{n=1}^{\infty} a_n.
  2. It converges to a sum that is a1a_1 less than the sum of ∑n=1∞an\sum_{n=1}^{\infty} a_n. (correct answer)
  3. It diverges because the starting term is different.
  4. Its convergence cannot be determined from the given information.
Explanation: The convergence or divergence of a series is not affected by removing a finite number of terms. Since ∑an\sum a_n converges, ∑an+1\sum a_{n+1} must also converge. Let S=∑n=1∞an=a1+a2+a3+…S = \sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \dots. The new series is ∑n=1∞an+1=a2+a3+a4+…\sum_{n=1}^{\infty} a_{n+1} = a_2 + a_3 + a_4 + \dots. This is the original series with the first term, a1a_1, removed. Therefore, its sum is S−a1S - a_1.

Question 20

Consider the series ∑n=1∞(1−2n)n\sum_{n=1}^{\infty} \left(1 - \frac{2}{n}\right)^n. Which of the following statements is true?

  1. The series converges because the limit of its terms is 1.
  2. The series converges to e−2e^{-2}.
  3. The series diverges because the limit of its terms is e−2≠0e^{-2} \neq 0. (correct answer)
  4. The n-th Term Test is inconclusive for this series.
Explanation: The n-th Term Test for Divergence states that if lim⁡n→∞an≠0\lim_{n \to \infty} a_n \neq 0, the series ∑an\sum a_n diverges. For this series, an=(1−2n)na_n = \left(1 - \frac{2}{n}\right)^n. The limit is a known form: lim⁡n→∞(1+kn)n=ek\lim_{n \to \infty} \left(1 + \frac{k}{n}\right)^n = e^k. Here, k=−2k = -2, so lim⁡n→∞an=e−2\lim_{n \to \infty} a_n = e^{-2}. Since e−2≠0e^{-2} \neq 0, the series diverges by the n-th Term Test.