Calculus 2 Quiz: Comparison Tests
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Comparison TestsQuestion 1 of 20

The Limit Comparison Test is applied to the series ∑n=1∞an=∑n=1∞ln⁡nn2\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{\ln n}{n^2} with the comparison series ∑n=1∞bn=∑n=1∞1n3/2\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}. What is the resulting limit and conclusion?

The limit is 0, and therefore ∑an\sum a_n converges.
The limit is 0, and the test is inconclusive.
The limit is ∞\infty, and therefore ∑an\sum a_n diverges.
The limit is 1, and therefore ∑an\sum a_n converges.
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Calculus 2 Quiz

Calculus 2 Quiz: Comparison Tests

Practice Comparison Tests 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 Comparison Tests, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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

The Limit Comparison Test is applied to the series ∑n=1∞an=∑n=1∞ln⁡nn2\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{\ln n}{n^2} with the comparison series ∑n=1∞bn=∑n=1∞1n3/2\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}. What is the resulting limit and conclusion?

  1. The limit is 0, and therefore ∑an\sum a_n converges. (correct answer)
  2. The limit is 0, and the test is inconclusive.
  3. The limit is ∞\infty, and therefore ∑an\sum a_n diverges.
  4. The limit is 1, and therefore ∑an\sum a_n converges.
Explanation: We compute the limit of the ratio: L=lim⁡n→∞anbn=lim⁡n→∞ln⁡(n)/n21/n3/2=lim⁡n→∞n3/2ln⁡nn2=lim⁡n→∞ln⁡nn1/2L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\ln(n)/n^2}{1/n^{3/2}} = \lim_{n \to \infty} \frac{n^{3/2} \ln n}{n^2} = \lim_{n \to \infty} \frac{\ln n}{n^{1/2}}. This is an indeterminate form ∞/∞\infty/\infty, so we can use L'Hôpital's Rule: L=lim⁡n→∞1/n(1/2)n−1/2=lim⁡n→∞2n=0L = \lim_{n \to \infty} \frac{1/n}{(1/2)n^{-1/2}} = \lim_{n \to \infty} \frac{2}{\sqrt{n}} = 0. In the Limit Comparison Test, when the limit L=0L=0 and the comparison series ∑bn\sum b_n converges, the original series ∑an\sum a_n must also converge. Here, ∑bn=∑1/n3/2\sum b_n = \sum 1/n^{3/2} is a convergent p-series (p=3/2>1p=3/2 > 1). Therefore, ∑an\sum a_n converges.

Question 2

Let ∑an=∑n=1∞3nn3\sum a_n = \sum_{n=1}^{\infty} \frac{3^n}{n^3}. Using the Limit Comparison Test with the series ∑bn=∑n=1∞1n\sum b_n = \sum_{n=1}^{\infty} \frac{1}{n}, what is the limit L=lim⁡n→∞anbnL = \lim_{n \to \infty} \frac{a_n}{b_n} and the correct conclusion?

  1. L=∞L = \infty, and the series ∑an\sum a_n diverges. (correct answer)
  2. L=∞L = \infty, and the test is inconclusive.
  3. L=0L = 0, and the test is inconclusive.
  4. L=3L = 3, and the series ∑an\sum a_n diverges.
Explanation: We are asked to compute L=lim⁡n→∞anbn=lim⁡n→∞3n/n31/n=lim⁡n→∞n⋅3nn3=lim⁡n→∞3nn2L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{3^n/n^3}{1/n} = \lim_{n \to \infty} \frac{n \cdot 3^n}{n^3} = \lim_{n \to \infty} \frac{3^n}{n^2}. Since exponential functions grow faster than polynomial functions, this limit is ∞\infty. According to the Limit Comparison Test, if L=∞L = \infty and the comparison series ∑bn\sum b_n diverges, then the original series ∑an\sum a_n also diverges. Here, ∑bn=∑1n\sum b_n = \sum \frac{1}{n} is the divergent harmonic series. Therefore, ∑an\sum a_n diverges. Note that the series also diverges by the Test for Divergence, as lim⁡n→∞an=∞\lim_{n \to \infty} a_n = \infty.

Question 3

Suppose an>0a_n > 0, bn>0b_n > 0, and lim⁡n→∞anbn=L\lim_{n \to \infty} \frac{a_n}{b_n} = L. Which combination of LL and the behavior of ∑bn\sum b_n is insufficient to determine the behavior of ∑an\sum a_n?

  1. L=0.5L = 0.5 and ∑bn\sum b_n converges.
  2. L=∞L = \infty and ∑bn\sum b_n diverges.
  3. L=0L = 0 and ∑bn\sum b_n converges.
  4. L=∞L = \infty and ∑bn\sum b_n converges. (correct answer)
Explanation: This question tests the edge cases of the Limit Comparison Test. (A) If 0<L<∞0 < L < \infty, ∑an\sum a_n and ∑bn\sum b_n have the same behavior. So if ∑bn\sum b_n converges, ∑an\sum a_n converges. Sufficient. (B) If L=∞L = \infty, ana_n is much larger than bnb_n. If the 'smaller' series ∑bn\sum b_n diverges, the 'larger' series ∑an\sum a_n must also diverge. Sufficient. (C) If L=0L = 0, ana_n is much smaller than bnb_n. If the 'larger' series ∑bn\sum b_n converges, the 'smaller' series ∑an\sum a_n must also converge. Sufficient. (D) If L=∞L = \infty, ana_n is much larger than bnb_n. If the 'smaller' series ∑bn\sum b_n converges, we can make no conclusion about the 'larger' series ∑an\sum a_n. It could converge or diverge. For example, let bn=1/n2b_n = 1/n^2 (converges). If an=1/na_n = 1/n, L=∞L=\infty and ∑an\sum a_n diverges. If an=1/n3/2a_n = 1/n^{3/2}, L=∞L=\infty and ∑an\sum a_n converges. Insufficient.

Question 4

Let ∑an\sum a_n be a series with positive terms. A student uses the Limit Comparison Test with ∑bn=∑1n\sum b_n = \sum \frac{1}{n} and finds lim⁡n→∞anbn=2\lim_{n \to \infty} \frac{a_n}{b_n} = 2. Then, the student uses the LCT with ∑cn=∑1n2\sum c_n = \sum \frac{1}{n^2} and finds lim⁡n→∞ancn=∞\lim_{n \to \infty} \frac{a_n}{c_n} = \infty. What can be concluded about ∑an\sum a_n?

  1. The series ∑an\sum a_n converges because the comparison with ∑cn\sum c_n is more definitive.
  2. The series ∑an\sum a_n diverges based on the comparison with ∑bn\sum b_n. (correct answer)
  3. The results are contradictory, so no conclusion can be drawn about ∑an\sum a_n.
  4. The series ∑an\sum a_n converges because lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0.
Explanation: The two tests provide consistent, not contradictory, information. The first test compares ∑an\sum a_n with the divergent harmonic series ∑bn=∑1/n\sum b_n = \sum 1/n. The limit of the ratio is 2, which is a finite, positive number. The LCT states that in this case, both series must have the same behavior. Since ∑1/n\sum 1/n diverges, ∑an\sum a_n must also diverge. The second test compares ∑an\sum a_n with the convergent p-series ∑cn=∑1/n2\sum c_n = \sum 1/n^2. The limit of the ratio is ∞\infty. The LCT states that if the limit is ∞\infty and the comparison series converges, the test is inconclusive. Therefore, the first test gives a definitive conclusion of divergence, while the second is inconclusive. The conclusion is that the series diverges.

Question 5

The limit comparison test is applied to ∑n=1∞n3+2nn2+sin⁡(n)\sum_{n=1}^{\infty} \frac{\sqrt{n^3 + 2n}}{n^2 + \sin(n)} using the comparison series ∑n=1∞1n\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}. What is the result of this test?

  1. The limit is 0, so the test is inconclusive about convergence of the original series
  2. The limit is 1, so both series have the same convergence behavior, and both diverge (correct answer)
  3. The limit is infinite, so the test is inconclusive about convergence of the original series
  4. The limit is 1, so both series have the same convergence behavior, and both converge
Explanation: For the limit comparison test: lim⁡n→∞n3+2nn2+sin⁡(n)1n=lim⁡n→∞n3+2n⋅nn2+sin⁡(n)=lim⁡n→∞n4+2n2n2+sin⁡(n)=lim⁡n→∞n21+2n2n2(1+sin⁡(n)n2)=1\lim_{n \to \infty} \frac{\frac{\sqrt{n^3 + 2n}}{n^2 + \sin(n)}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{\sqrt{n^3 + 2n} \cdot \sqrt{n}}{n^2 + \sin(n)} = \lim_{n \to \infty} \frac{\sqrt{n^4 + 2n^2}}{n^2 + \sin(n)} = \lim_{n \to \infty} \frac{n^2\sqrt{1 + \frac{2}{n^2}}}{n^2(1 + \frac{\sin(n)}{n^2})} = 1. Since the limit is 1 (positive and finite), both series have the same convergence behavior. The comparison series ∑1n=∑1n1/2\sum \frac{1}{\sqrt{n}} = \sum \frac{1}{n^{1/2}} is a p-series with p=12<1p = \frac{1}{2} < 1, so it diverges. Therefore, the original series also diverges.

Question 6

Consider applying the limit comparison test to ∑n=1∞an\sum_{n=1}^{\infty} a_n and ∑n=1∞bn\sum_{n=1}^{\infty} b_n where an,bn>0a_n, b_n > 0. If lim⁡n→∞anbn=0\lim_{n \to \infty} \frac{a_n}{b_n} = 0 and ∑bn\sum b_n converges, what additional information is needed to determine the convergence of ∑an\sum a_n?

  1. The limit comparison test is inconclusive when the limit is 0; we need a different convergence test
  2. We need to verify that ana_n approaches zero faster than bnb_n for the test to be conclusive
  3. We need to know whether ∑an\sum a_n satisfies the conditions for direct comparison with ∑bn\sum b_n
  4. No additional information is needed; ∑an\sum a_n must converge by the limit comparison test (correct answer)
Explanation: When you encounter limit comparison test problems, remember that this test has three possible outcomes based on the limit value, and each tells you something different about convergence behavior. The limit comparison test states that if lim⁡n→∞anbn=L\lim_{n \to \infty} \frac{a_n}{b_n} = L, then: if L>0L > 0 and finite, both series have the same convergence behavior; if L=0L = 0 and ∑bn\sum b_n converges, then ∑an\sum a_n also converges; if L=∞L = \infty and ∑bn\sum b_n diverges, then ∑an\sum a_n also diverges. In this problem, we have lim⁡n→∞anbn=0\lim_{n \to \infty} \frac{a_n}{b_n} = 0 and ∑bn\sum b_n converges. This fits the second case perfectly: when the limit is 0 and the comparison series converges, the original series must also converge. The intuition is that ana_n approaches zero faster than bnb_n, so if the "larger" series ∑bn\sum b_n converges, the "smaller" series ∑an\sum a_n must also converge. Choice A incorrectly claims the test is inconclusive when the limit is 0—this is false. Choice B suggests we need additional verification about the rate of approach to zero, but the limit calculation already establishes this. Choice C mentions needing conditions for direct comparison, but limit comparison is designed to work when direct comparison might fail. Remember: limit comparison test conclusions depend entirely on the limit value and the convergence status of your comparison series. When L=0L = 0 with a convergent comparison series, you immediately know convergence—no additional work required.

Question 7

Consider ∑n=2∞1n(ln⁡n)p\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p} where p>0p > 0. Using the integral test as a comparison tool, for which values of p does this series converge?

  1. p>1p > 1 only, because the corresponding improper integral converges precisely when p>1p > 1 (correct answer)
  2. p≥1p \geq 1 only, because we need the integral to converge at the boundary case
  3. p>0p > 0 for all positive values, because the logarithmic factor always ensures convergence
  4. p≥2p \geq 2 only, because we need stronger decay than the harmonic series provides
Explanation: The integral ∫2∞1x(ln⁡x)pdx\int_2^{\infty} \frac{1}{x(\ln x)^p} dx uses substitution u=ln⁡xu = \ln x, giving ∫ln⁡2∞1updu\int_{\ln 2}^{\infty} \frac{1}{u^p} du. This converges if and only if p>1p > 1. When p=1p = 1, we get ∫1udu=ln⁡u\int \frac{1}{u} du = \ln u, which diverges. When p<1p < 1, the integral also diverges. When p>1p > 1, we get u1−p1−p\frac{u^{1-p}}{1-p} which converges. Therefore, the series converges exactly when p>1p > 1.

Question 8

Which of the following determines the convergence of the series ∑n=1∞sin⁡(1n2)\sum_{n=1}^{\infty} \sin\left(\frac{1}{n^2}\right)?

  1. The series converges by the Limit Comparison Test with ∑1n2\sum \frac{1}{n^2}, because lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1. (correct answer)
  2. The series diverges by the Limit Comparison Test with ∑1n\sum \frac{1}{n}, because sin⁡(x)\sin(x) behaves like xx.
  3. The series converges because lim⁡n→∞sin⁡(1/n2)=0\lim_{n \to \infty} \sin(1/n^2) = 0, which is a sufficient condition for convergence.
  4. Comparison tests cannot be applied because the terms sin⁡(1/n2)\sin(1/n^2) are not guaranteed to be positive for all nn.
Explanation: For the series ∑an=∑sin⁡(1/n2)\sum a_n = \sum \sin(1/n^2), we use the Limit Comparison Test. A suitable comparison series is ∑bn=∑1n2\sum b_n = \sum \frac{1}{n^2}, which is a convergent p-series. We evaluate the limit lim⁡n→∞anbn=lim⁡n→∞sin⁡(1/n2)1/n2\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\sin(1/n^2)}{1/n^2}. Let x=1/n2x = 1/n^2; as n→∞n \to \infty, x→0x \to 0. The limit becomes lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1. Since the limit is a finite, positive number, and ∑1n2\sum \frac{1}{n^2} converges, the original series also converges.

Question 9

To determine the convergence of the series ∑n=2∞1(ln⁡n)4\sum_{n=2}^{\infty} \frac{1}{(\ln n)^4}, which comparison demonstrates divergence using the Direct Comparison Test?

  1. For large nn, 1(ln⁡n)4>1n\frac{1}{(\ln n)^4} > \frac{1}{n}, and ∑1n\sum \frac{1}{n} diverges. (correct answer)
  2. For large nn, 1(ln⁡n)4<1ln⁡n\frac{1}{(\ln n)^4} < \frac{1}{\ln n}, and ∑1ln⁡n\sum \frac{1}{\ln n} diverges.
  3. For large nn, 1(ln⁡n)4>1n2\frac{1}{(\ln n)^4} > \frac{1}{n^2}, and ∑1n2\sum \frac{1}{n^2} converges.
  4. For large nn, 1(ln⁡n)4>1en\frac{1}{(\ln n)^4} > \frac{1}{e^n}, and ∑1en\sum \frac{1}{e^n} converges.
Explanation: To prove divergence via the Direct Comparison Test, we need to show that the series' terms are greater than or equal to the terms of a known divergent series. We know that for any positive power kk, n>(ln⁡n)kn > (\ln n)^k for sufficiently large nn. In this case, for large nn, we have n>(ln⁡n)4n > (\ln n)^4. Taking the reciprocal reverses the inequality: 1n<1(ln⁡n)4\frac{1}{n} < \frac{1}{(\ln n)^4}. Since ∑1n\sum \frac{1}{n} (the harmonic series) diverges, and our terms are larger, the series ∑1(ln⁡n)4\sum \frac{1}{(\ln n)^4} must also diverge by the Direct Comparison Test.

Question 10

Which of the following statements correctly determines the convergence of ∑n=2∞nn−1\sum_{n=2}^{\infty} \frac{\sqrt{n}}{n-1}?

  1. The series diverges by Direct Comparison, as nn−1>nn=1n\frac{\sqrt{n}}{n-1} > \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}} and ∑1n\sum \frac{1}{\sqrt{n}} diverges. (correct answer)
  2. The series converges by Direct Comparison, as nn−1<2n\frac{\sqrt{n}}{n-1} < \frac{2}{\sqrt{n}} for n≥2n \ge 2, and ∑2n\sum \frac{2}{\sqrt{n}} converges.
  3. The series converges by Limit Comparison with ∑1n3/2\sum \frac{1}{n^{3/2}}, because the terms have similar algebraic form.
  4. The series converges because the limit of its terms approaches zero.
Explanation: The terms of the series, an=nn−1a_n = \frac{\sqrt{n}}{n-1}, behave like nn=1n1/2\frac{\sqrt{n}}{n} = \frac{1}{n^{1/2}}. The series ∑1n\sum \frac{1}{\sqrt{n}} is a divergent p-series (p=1/2p=1/2). We can use the Direct Comparison Test. For n≥2n \ge 2, we have n−1<nn-1 < n. Taking the reciprocal reverses the inequality: 1n−1>1n\frac{1}{n-1} > \frac{1}{n}. Multiplying by n\sqrt{n} (a positive number) preserves the inequality: nn−1>nn=1n\frac{\sqrt{n}}{n-1} > \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}}. Since the terms of our series are greater than the terms of a known divergent series, the series ∑nn−1\sum \frac{\sqrt{n}}{n-1} diverges.

Question 11

Consider the series ∑n=1∞an\sum_{n=1}^{\infty} a_n, where an=2n2−n+1n4−n2+5a_n = \frac{2n^2 - n + 1}{n^4 - n^2 + 5}. A valid conclusion using the Limit Comparison Test is that the series:

  1. converges, by comparing with ∑1n2\sum \frac{1}{n^2} because the limit of the ratio of terms is 2. (correct answer)
  2. diverges, by comparing with ∑1n\sum \frac{1}{n} because the terms are always positive.
  3. converges, by comparing with ∑1n4\sum \frac{1}{n^4} because the limit of the ratio of terms is 1.
  4. diverges, by comparing with ∑1n2\sum \frac{1}{n^2} because the limit of the ratio of terms is 2.
Explanation: For large nn, the term an=2n2−n+1n4−n2+5a_n = \frac{2n^2 - n + 1}{n^4 - n^2 + 5} behaves like 2n2n4=2n2\frac{2n^2}{n^4} = \frac{2}{n^2}. This suggests using the Limit Comparison Test with bn=1n2b_n = \frac{1}{n^2}. The limit of the ratio is lim⁡n→∞anbn=lim⁡n→∞(2n2−n+1)/(n4−n2+5)1/n2=lim⁡n→∞n2(2n2−n+1)n4−n2+5=lim⁡n→∞2n4−n3+n2n4−n2+5=2\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{(2n^2 - n + 1)/(n^4 - n^2 + 5)}{1/n^2} = \lim_{n \to \infty} \frac{n^2(2n^2 - n + 1)}{n^4 - n^2 + 5} = \lim_{n \to \infty} \frac{2n^4 - n^3 + n^2}{n^4 - n^2 + 5} = 2. Since the limit is 2 (a finite, positive number) and the comparison series ∑1n2\sum \frac{1}{n^2} converges (p-series with p=2>1p=2>1), the original series ∑an\sum a_n also converges.

Question 12

If ∑n=1∞an\sum_{n=1}^{\infty} a_n is a convergent series with positive terms, which of the following series is guaranteed to converge?

  1. ∑n=1∞1an\sum_{n=1}^{\infty} \frac{1}{a_n}
  2. ∑n=1∞an\sum_{n=1}^{\infty} \sqrt{a_n}
  3. ∑n=1∞an2\sum_{n=1}^{\infty} a_n^2 (correct answer)
  4. ∑n=1∞nan\sum_{n=1}^{\infty} n a_n
Explanation: If ∑an\sum a_n converges, then lim⁡n→∞an=0\lim_{n \to \infty} a_n = 0. This means for nn large enough, 0<an<10 < a_n < 1. When 0<x<10 < x < 1, we have x2<xx^2 < x. Therefore, for sufficiently large nn, 0<an2<an0 < a_n^2 < a_n. Since ∑an\sum a_n converges, by the Direct Comparison Test, ∑an2\sum a_n^2 must also converge. For the other options: (A) If an→0a_n \to 0, then 1/an→∞1/a_n \to \infty, so ∑1/an\sum 1/a_n diverges. (B) Taking an=1/n2a_n = 1/n^2, we have ∑an\sum a_n converges, but ∑an=∑1/n\sum \sqrt{a_n} = \sum 1/n diverges. (D) Taking an=1/n2a_n = 1/n^2, we get ∑nan=∑1/n\sum na_n = \sum 1/n which diverges.

Question 13

Consider the series ∑n=1∞12n−n\sum_{n=1}^{\infty} \frac{1}{2^n - n}. Which application of a comparison test is valid?

  1. Diverges by Direct Comparison, since 12n−n>12n\frac{1}{2^n - n} > \frac{1}{2^n} and ∑12n\sum \frac{1}{2^n} converges.
  2. Converges by Limit Comparison with ∑12n\sum \frac{1}{2^n}, as the limit of the ratio is 1. (correct answer)
  3. Converges by Direct Comparison, since 12n−n<1n\frac{1}{2^n - n} < \frac{1}{n} and ∑1n\sum \frac{1}{n} diverges.
  4. Diverges by Limit Comparison with ∑1n\sum \frac{1}{n}, as the limit of the ratio is ∞\infty.
Explanation: The dominant term in the denominator is 2n2^n. We compare with the convergent geometric series ∑bn=∑12n\sum b_n = \sum \frac{1}{2^n}. Using Direct Comparison, 2n−n<2n2^n - n < 2^n implies 12n−n>12n\frac{1}{2^n - n} > \frac{1}{2^n}. This inequality goes the wrong way to prove convergence. So we use the Limit Comparison Test. L=lim⁡n→∞1/(2n−n)1/2n=lim⁡n→∞2n2n−n=lim⁡n→∞11−n/2nL = \lim_{n \to \infty} \frac{1/(2^n - n)}{1/2^n} = \lim_{n \to \infty} \frac{2^n}{2^n - n} = \lim_{n \to \infty} \frac{1}{1 - n/2^n}. Since exponential functions grow faster than polynomials, lim⁡n→∞n/2n=0\lim_{n \to \infty} n/2^n = 0. Therefore, L=1L=1. Since LL is a finite positive number and ∑1/2n\sum 1/2^n converges, the original series converges.

Question 14

Let ∑an\sum a_n and ∑bn\sum b_n be series with positive terms. If lim⁡n→∞anbn=0\lim_{n \to \infty} \frac{a_n}{b_n} = 0 and ∑bn\sum b_n diverges, what can be concluded about ∑an\sum a_n?

  1. The series ∑an\sum a_n converges.
  2. The series ∑an\sum a_n diverges.
  3. The series ∑an\sum a_n converges only if an<bna_n < b_n for all nn.
  4. No conclusion can be drawn without more information. (correct answer)
Explanation: This scenario represents an inconclusive case of the Limit Comparison Test. When lim⁡n→∞anbn=0\lim_{n \to \infty} \frac{a_n}{b_n} = 0, it means that for large nn, ana_n is significantly smaller than bnb_n. If ∑bn\sum b_n were to converge, we could conclude that ∑an\sum a_n converges. However, since ∑bn\sum b_n diverges, knowing that ∑an\sum a_n is 'smaller' provides no information. For example, if bn=1/nb_n = 1/n (diverges), and an=1/n2a_n = 1/n^2, then lim⁡anbn=0\lim \frac{a_n}{b_n} = 0 and ∑an\sum a_n converges. But if bn=1/nb_n = 1/n and an=1/(nln⁡n)a_n = 1/(n \ln n) (for n≥2n \ge 2), then lim⁡anbn=0\lim \frac{a_n}{b_n} = 0 and ∑an\sum a_n diverges. Thus, no conclusion can be drawn.

Question 15

What is the behavior of the series ∑n=1∞1n1+1/n\sum_{n=1}^{\infty} \frac{1}{n^{1+1/n}}?

  1. Converges, because the exponent 1+1/n1+1/n is strictly greater than 1 for all nn.
  2. Diverges, by Limit Comparison with the harmonic series ∑1n\sum \frac{1}{n}. (correct answer)
  3. Converges, by Limit Comparison with the p-series ∑1n2\sum \frac{1}{n^2}.
  4. Diverges, because lim⁡n→∞1n1+1/n≠0\lim_{n \to \infty} \frac{1}{n^{1+1/n}} ≠ 0.
Explanation: The exponent 1+1/n1+1/n approaches 1 as n→∞n \to \infty, so the term an=1n1+1/na_n = \frac{1}{n^{1+1/n}} behaves like 1n\frac{1}{n}. This suggests using the Limit Comparison Test with the harmonic series ∑bn=∑1n\sum b_n = \sum \frac{1}{n}. We compute the limit of the ratio: L=lim⁡n→∞anbn=lim⁡n→∞1/n1+1/n1/n=lim⁡n→∞nn⋅n1/n=lim⁡n→∞1n1/nL = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{1/n^{1+1/n}}{1/n} = \lim_{n \to \infty} \frac{n}{n \cdot n^{1/n}} = \lim_{n \to \infty} \frac{1}{n^{1/n}}. Using the known limit lim⁡n→∞n1/n=1\lim_{n \to \infty} n^{1/n} = 1, we find L=1L=1. Since LL is a finite, positive number and the comparison series ∑1n\sum \frac{1}{n} diverges, the original series also diverges.

Question 16

The series ∑n=1∞n(n+1)3n\sum_{n=1}^{\infty} \frac{n}{(n+1)3^n} is being tested for convergence. Which statement is correct?

  1. The series diverges by Limit Comparison with ∑13n\sum \frac{1}{3^n}.
  2. The series converges by Direct Comparison with ∑13n\sum \frac{1}{3^n}. (correct answer)
  3. The series diverges by Direct Comparison with ∑1n\sum \frac{1}{n}.
  4. The series converges by Limit Comparison with ∑1n\sum \frac{1}{n}.
Explanation: Let an=n(n+1)3na_n = \frac{n}{(n+1)3^n}. We can use the Direct Comparison Test. For all n≥1n \ge 1, the fraction nn+1\frac{n}{n+1} is less than 1. Therefore, we can write the inequality: an=nn+1⋅13n<1⋅13n=13na_n = \frac{n}{n+1} \cdot \frac{1}{3^n} < 1 \cdot \frac{1}{3^n} = \frac{1}{3^n}. Let bn=13n=(13)nb_n = \frac{1}{3^n} = (\frac{1}{3})^n. The series ∑bn\sum b_n is a geometric series with ratio r=1/3r = 1/3. Since ∣r∣<1|r| < 1, ∑bn\sum b_n converges. Because 0<an<bn0 < a_n < b_n and ∑bn\sum b_n converges, the original series ∑an\sum a_n must also converge by the Direct Comparison Test.

Question 17

What is the behavior of the series ∑n=1∞5+cos⁡(nπ)nn\sum_{n=1}^{\infty} \frac{5 + \cos(n\pi)}{n\sqrt{n}}?

  1. It diverges, because the term cos⁡(nπ)\cos(n\pi) oscillates between -1 and 1.
  2. It is a conditionally convergent alternating series.
  3. It converges, by Direct Comparison with ∑6n3/2\sum \frac{6}{n^{3/2}}. (correct answer)
  4. It diverges, by Direct Comparison with ∑4n3/2\sum \frac{4}{n^{3/2}}.
Explanation: The term cos⁡(nπ)\cos(n\pi) is equal to (−1)n(-1)^n. Thus, the numerator 5+cos⁡(nπ)5 + \cos(n\pi) alternates between 5−1=45-1=4 and 5+1=65+1=6. Since the numerator is always positive, all terms of the series are positive, and comparison tests apply. We can bound the general term ana_n: 4nn≤an=5+cos⁡(nπ)nn≤6nn\frac{4}{n\sqrt{n}} \le a_n = \frac{5 + \cos(n\pi)}{n\sqrt{n}} \le \frac{6}{n\sqrt{n}}. Let's use the upper bound for a Direct Comparison. We compare our series with ∑bn=∑6nn=∑6n3/2\sum b_n = \sum \frac{6}{n\sqrt{n}} = \sum \frac{6}{n^{3/2}}. This is a constant multiple of a p-series with p=3/2>1p=3/2 > 1, so it converges. Since 0≤an≤bn0 \le a_n \le b_n and ∑bn\sum b_n converges, the original series must also converge.

Question 18

Which of the following series can be shown to diverge using the Direct Comparison Test with the harmonic series ∑n=3∞1n\sum_{n=3}^{\infty} \frac{1}{n}?

  1. ∑n=3∞1n2−4\sum_{n=3}^{\infty} \frac{1}{n^2 - 4}
  2. ∑n=3∞ln⁡nn\sum_{n=3}^{\infty} \frac{\ln n}{n} (correct answer)
  3. ∑n=3∞nn3+1\sum_{n=3}^{\infty} \frac{n}{n^3 + 1}
  4. ∑n=3∞1n+ln⁡n\sum_{n=3}^{\infty} \frac{1}{n+ \ln n}
Explanation: To prove divergence by Direct Comparison with ∑1n\sum \frac{1}{n}, we must find a series ∑an\sum a_n such that an≥1na_n \ge \frac{1}{n} for sufficiently large nn. (A) 1n2−4\frac{1}{n^2 - 4} behaves like 1n2\frac{1}{n^2}, not 1n\frac{1}{n}. (B) For n≥3n \ge 3, ln⁡n>ln⁡(e)=1\ln n > \ln(e) = 1. Therefore, ln⁡nn>1n\frac{\ln n}{n} > \frac{1}{n}. Since ∑1n\sum \frac{1}{n} diverges, ∑ln⁡nn\sum \frac{\ln n}{n} must also diverge by the Direct Comparison Test. (C) nn3+1<nn3=1n2\frac{n}{n^3 + 1} < \frac{n}{n^3} = \frac{1}{n^2}, and comparison with 1n\frac{1}{n} is not straightforward for divergence. (D) n+ln⁡n>nn + \ln n > n, so 1n+ln⁡n<1n\frac{1}{n + \ln n} < \frac{1}{n}. This inequality goes the wrong way to prove divergence by comparing with ∑1n\sum \frac{1}{n}.

Question 19

Consider the series ∑n=1∞n!(n+2)!\sum_{n=1}^{\infty} \frac{n!}{(n+2)!}. Which statement correctly describes the application of a comparison test?

  1. The series diverges by Limit Comparison with ∑1n\sum \frac{1}{n}, as the limit of the ratio of terms is 1.
  2. The series diverges by Direct Comparison, since n!(n+2)!>1n3\frac{n!}{(n+2)!} > \frac{1}{n^3} and ∑1n3\sum \frac{1}{n^3} converges.
  3. The series converges by Direct Comparison with ∑1n2\sum \frac{1}{n^2}, since n!(n+2)!<1n2\frac{n!}{(n+2)!} < \frac{1}{n^2}. (correct answer)
  4. The series converges by Limit Comparison with ∑1n\sum \frac{1}{n}, as the limit of the ratio of terms is 0.
Explanation: First, simplify the general term: an=n!(n+2)!=n!(n+2)(n+1)n!=1(n+2)(n+1)=1n2+3n+2a_n = \frac{n!}{(n+2)!} = \frac{n!}{(n+2)(n+1)n!} = \frac{1}{(n+2)(n+1)} = \frac{1}{n^2+3n+2}. This term behaves like 1n2\frac{1}{n^2} for large nn. Let's use ∑bn=∑1n2\sum b_n = \sum \frac{1}{n^2} as the comparison series, which is a convergent p-series. For the Direct Comparison Test, we check the inequality: n2+3n+2>n2n^2+3n+2 > n^2 for n≥1n \ge 1, which implies 1n2+3n+2<1n2\frac{1}{n^2+3n+2} < \frac{1}{n^2}. Since an<bna_n < b_n and ∑bn\sum b_n converges, the original series converges by the Direct Comparison Test.

Question 20

Let an=n3+4nn!a_n = \frac{n^3 + 4^n}{n!}. Which of the following is the most effective way to prove the convergence of ∑an\sum a_n using a comparison test?

  1. Use Limit Comparison with ∑n3n!\sum \frac{n^3}{n!}, which converges.
  2. Use Direct Comparison with ∑2⋅n3n!\sum \frac{2 \cdot n^3}{n!} for large nn, which converges.
  3. Use Limit Comparison with ∑4nn!\sum \frac{4^n}{n!}, which converges. (correct answer)
  4. Use Direct Comparison with ∑1n4\sum \frac{1}{n^4}, which converges.
Explanation: The term is an=n3+4nn!a_n = \frac{n^3 + 4^n}{n!}. For large nn, the exponential term 4n4^n dominates the polynomial term n3n^3 in the numerator. Therefore, the most natural and effective comparison series is one that captures this dominant behavior, which is ∑bn=∑4nn!\sum b_n = \sum \frac{4^n}{n!}. The convergence of ∑bn\sum b_n can be established by the Ratio Test (lim⁡n→∞4n+1=0<1\lim_{n \to \infty} \frac{4}{n+1} = 0 < 1). Using the Limit Comparison Test on ∑an\sum a_n with ∑bn\sum b_n: lim⁡n→∞anbn=lim⁡n→∞(n3+4n)/n!(4n)/n!=lim⁡n→∞n3+4n4n=lim⁡n→∞(n34n+1)=0+1=1\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{(n^3+4^n)/n!}{(4^n)/n!} = \lim_{n \to \infty} \frac{n^3+4^n}{4^n} = \lim_{n \to \infty} (\frac{n^3}{4^n} + 1) = 0 + 1 = 1. Since the limit is 1 (a finite, positive number) and the comparison series converges, ∑an\sum a_n converges.