Comparison Tests - Calculus 2

Given that $\sum_{n=1}^{\infty} a_n$ converges and $a_n > 0$ for all n, consider the series $\sum_{n=1}^{\infty} \frac{a_n}{1 + a_n}$ . What can be concluded about its convergence?
Answer: It converges by direct comparison since $\frac{a_n}{1 + a_n} < a_n$ and $\sum a_n$ converges Since $a_n > 0$ , we have $1 + a_n > 1$ , so $\frac{a_n}{1 + a_n} < \frac{a_n}{1} = a_n$ . Since $\sum a_n$ converges and $0 < \frac{a_n}{1 + a_n} < a_n$ , the direct comparison test guarantees that $\sum \frac{a_n}{1 + a_n}$ converges. Choice B incorrectly states the limit; since $\sum a_n$ converges, $a_n \to 0$ , so $\frac{a_n}{1 + a_n} \to 0$ . Choice C has the right idea but the limit is actually 1, making it equivalent to choice A but less direct.
Consider the series $\sum_{n=1}^{\infty} \frac{1}{2^n - n}$ . Which application of a comparison test is valid?
Answer: Converges by Limit Comparison with $\sum \frac{1}{2^n}$ , as the limit of the ratio is 1. The dominant term in the denominator is $2^n$ . We compare with the convergent geometric series $\sum b_n = \sum \frac{1}{2^n}$ . Using Direct Comparison, $2^n - n < 2^n$ implies $\frac{1}{2^n - n} > \frac{1}{2^n}$ . This inequality goes the wrong way to prove convergence. So we use the Limit Comparison Test. $L = \lim_{n \to \infty} \frac{1/(2^n - n)}{1/2^n} = \lim_{n \to \infty} \frac{2^n}{2^n - n} = \lim_{n \to \infty} \frac{1}{1 - n/2^n}$ . Since exponential functions grow faster than polynomials, $\lim_{n \to \infty} n/2^n = 0$ . Therefore, $L=1$ . Since $L$ is a finite positive number and $\sum 1/2^n$ converges, the original series converges.
What is the behavior of the series $\sum_{n=1}^{\infty} \frac{1}{n^{1+1/n}}$ ?
Answer: Diverges, by Limit Comparison with the harmonic series $\sum \frac{1}{n}$ . The exponent $1+1/n$ approaches 1 as $n \to \infty$ , so the term $a_n = \frac{1}{n^{1+1/n}}$ behaves like $\frac{1}{n}$ . This suggests using the Limit Comparison Test with the harmonic series $\sum b_n = \sum \frac{1}{n}$ . We compute the limit of the ratio: $L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{1/n^{1+1/n}}{1/n} = \lim_{n \to \infty} \frac{n}{n \cdot n^{1/n}} = \lim_{n \to \infty} \frac{1}{n^{1/n}}$ . Using the known limit $\lim_{n \to \infty} n^{1/n} = 1$ , we find $L=1$ . Since $L$ is a finite, positive number and the comparison series $\sum \frac{1}{n}$ diverges, the original series also diverges.
To investigate the series $\sum_{n=1}^{\infty} \frac{1}{n(1+\ln n)}$ , an appropriate comparison series $\sum b_n$ and test would be:
Answer: $\sum b_n = \sum \frac{1}{n \ln n}$ , which shows divergence, implying the original series also diverges. The term $a_n = \frac{1}{n(1+\ln n)}$ looks very similar to $b_n = \frac{1}{n \ln n}$ . The series $\sum_{n=2}^{\infty} \frac{1}{n \ln n}$ is a known divergent series (by the Integral Test). Let's use the Limit Comparison Test with $\sum a_n$ and $\sum b_n$ (starting at n=2 to avoid issues with ln(1)). $L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{1/(n(1+\ln n))}{1/(n \ln n)} = \lim_{n \to \infty} \frac{n \ln n}{n(1+\ln n)} = \lim_{n \to \infty} \frac{\ln n}{1+\ln n} = 1$ . Since $L=1$ (a finite, positive number) and the comparison series $\sum \frac{1}{n \ln n}$ diverges, the original series $\sum \frac{1}{n(1+\ln n)}$ must also diverge.
Let $\sum a_n$ be a series with positive terms. A student uses the Limit Comparison Test with $\sum b_n = \sum \frac{1}{n}$ and finds $\lim_{n \to \infty} \frac{a_n}{b_n} = 2$ . Then, the student uses the LCT with $\sum c_n = \sum \frac{1}{n^2}$ and finds $\lim_{n \to \infty} \frac{a_n}{c_n} = \infty$ . What can be concluded about $\sum a_n$ ?
Answer: The series $\sum a_n$ diverges based on the comparison with $\sum b_n$ . The two tests provide consistent, not contradictory, information. The first test compares $\sum a_n$ with the divergent harmonic series $\sum b_n = \sum 1/n$ . 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 $\sum 1/n$ diverges, $\sum a_n$ must also diverge. The second test compares $\sum a_n$ with the convergent p-series $\sum c_n = \sum 1/n^2$ . 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.
The Limit Comparison Test is applied to the series $\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{\ln n}{n^2}$ with the comparison series $\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ . What is the resulting limit and conclusion?
Answer: The limit is 0, and therefore $\sum a_n$ converges. We compute the limit of the ratio: $L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\ln(n)/n^2}{1/n^{3/2}} = \lim_{n \to \infty} \frac{n^{3/2} \ln n}{n^2} = \lim_{n \to \infty} \frac{\ln n}{n^{1/2}}$ . This is an indeterminate form $\infty/\infty$ , so we can use L'Hôpital's Rule: $L = \lim_{n \to \infty} \frac{1/n}{(1/2)n^{-1/2}} = \lim_{n \to \infty} \frac{2}{\sqrt{n}} = 0$ . In the Limit Comparison Test, when the limit $L=0$ and the comparison series $\sum b_n$ converges, the original series $\sum a_n$ must also converge. Here, $\sum b_n = \sum 1/n^{3/2}$ is a convergent p-series ( $p=3/2 > 1$ ). Therefore, $\sum a_n$ converges.
Let $a_n = \frac{n^3 + 4^n}{n!}$ . Which of the following is the most effective way to prove the convergence of $\sum a_n$ using a comparison test?
Answer: Use Limit Comparison with $\sum \frac{4^n}{n!}$ , which converges. The term is $a_n = \frac{n^3 + 4^n}{n!}$ . For large $n$ , the exponential term $4^n$ dominates the polynomial term $n^3$ in the numerator. Therefore, the most natural and effective comparison series is one that captures this dominant behavior, which is $\sum b_n = \sum \frac{4^n}{n!}$ . The convergence of $\sum b_n$ can be established by the Ratio Test ( $\lim_{n \to \infty} \frac{4}{n+1} = 0 < 1$ ). Using the Limit Comparison Test on $\sum a_n$ with $\sum b_n$ : $\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{(n^3+4^n)/n!}{(4^n)/n!} = \lim_{n \to \infty} \frac{n^3+4^n}{4^n} = \lim_{n \to \infty} (\frac{n^3}{4^n} + 1) = 0 + 1 = 1$ . Since the limit is 1 (a finite, positive number) and the comparison series converges, $\sum a_n$ converges.
Consider the series $\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (1 - \cos(\frac{1}{n}))$ . Which comparison leads to the correct conclusion?
Answer: Converges, by Limit Comparison with $\sum \frac{1}{n^2}$ . We analyze the term $a_n = 1 - \cos(1/n)$ . The Taylor expansion for $\cos(x)$ around $x=0$ is $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$ . So, for $x=1/n$ , $\cos(1/n) \approx 1 - \frac{1}{2n^2}$ . This means $a_n = 1 - \cos(1/n) \approx 1 - (1 - \frac{1}{2n^2}) = \frac{1}{2n^2}$ . This suggests comparison with $\sum b_n = \sum \frac{1}{n^2}$ . Using the Limit Comparison Test: $L = \lim_{n \to \infty} \frac{1 - \cos(1/n)}{1/n^2}$ . Let $x=1/n$ . The limit is $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$ . By L'Hôpital's Rule, this is $\lim_{x \to 0} \frac{\sin x}{2x} = \frac{1}{2}$ . Since $L=1/2$ is a finite positive number and $\sum \frac{1}{n^2}$ converges, the original series converges.
Consider the series $\sum_{n=1}^{\infty} \frac{\ln(n+2)}{n^2 + 3n + 1}$ . To determine convergence using the limit comparison test, which comparison series would be most effective?
Answer: $\sum_{n=1}^{\infty} \frac{\ln n}{n^2}$ because it preserves both the logarithmic and quadratic behaviors For large n, $\ln(n+2) \sim \ln n$ and $n^2 + 3n + 1 \sim n^2$ . The series $\sum \frac{\ln n}{n^2}$ captures both behaviors and converges by the integral test. Choice A uses $\sum \frac{1}{n}$ which diverges and doesn't account for the logarithmic growth. Choice C ignores the logarithmic term entirely. Choice D gives $\sum \frac{\ln n}{n}$ which diverges, making it unsuitable for comparison.
Which of the following statements correctly determines the convergence of $\sum_{n=2}^{\infty} \frac{\sqrt{n}}{n-1}$ ?
Answer: The series diverges by Direct Comparison, as $\frac{\sqrt{n}}{n-1} > \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}}$ and $\sum \frac{1}{\sqrt{n}}$ diverges. The terms of the series, $a_n = \frac{\sqrt{n}}{n-1}$ , behave like $\frac{\sqrt{n}}{n} = \frac{1}{n^{1/2}}$ . The series $\sum \frac{1}{\sqrt{n}}$ is a divergent p-series ( $p=1/2$ ). We can use the Direct Comparison Test. For $n \ge 2$ , we have $n-1 < n$ . Taking the reciprocal reverses the inequality: $\frac{1}{n-1} > \frac{1}{n}$ . Multiplying by $\sqrt{n}$ (a positive number) preserves the inequality: $\frac{\sqrt{n}}{n-1} > \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}}$ . Since the terms of our series are greater than the terms of a known divergent series, the series $\sum \frac{\sqrt{n}}{n-1}$ diverges.
Consider the series $\sum_{n=1}^{\infty} a_n$ , where $a_n = \frac{2n^2 - n + 1}{n^4 - n^2 + 5}$ . A valid conclusion using the Limit Comparison Test is that the series:
Answer: converges, by comparing with $\sum \frac{1}{n^2}$ because the limit of the ratio of terms is 2. For large $n$ , the term $a_n = \frac{2n^2 - n + 1}{n^4 - n^2 + 5}$ behaves like $\frac{2n^2}{n^4} = \frac{2}{n^2}$ . This suggests using the Limit Comparison Test with $b_n = \frac{1}{n^2}$ . The limit of the ratio is $\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{(2n^2 - n + 1)/(n^4 - n^2 + 5)}{1/n^2} = \lim_{n \to \infty} \frac{n^2(2n^2 - n + 1)}{n^4 - n^2 + 5} = \lim_{n \to \infty} \frac{2n^4 - n^3 + n^2}{n^4 - n^2 + 5} = 2$ . Since the limit is 2 (a finite, positive number) and the comparison series $\sum \frac{1}{n^2}$ converges (p-series with $p=2>1$ ), the original series $\sum a_n$ also converges.
For which of the following series would the direct comparison test be inconclusive when compared to $\sum_{n=1}^{\infty} \frac{1}{n}$ ?
Answer: $\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ because the terms are smaller than $\frac{1}{n}$ but $\sum \frac{1}{n}$ diverges The direct comparison test is inconclusive when we have $0 \leq a_n \leq b_n$ but $\sum b_n$ diverges - we cannot conclude anything about $\sum a_n$ . Choice A: $\frac{1}{n^2+1} < \frac{1}{n}$ and $\sum \frac{1}{n}$ diverges, making the test inconclusive. Choice B: $\frac{n+1}{n^2} > \frac{1}{n}$ for large n and $\sum \frac{1}{n}$ diverges, so by comparison the series diverges (conclusive). Choice C: $\frac{2n+1}{n^2} > \frac{1}{n}$ for large n, so the test concludes divergence. Choice D: $\frac{2n}{n^2} = \frac{2}{n}$ and since $\sum \frac{2}{n}$ diverges, this gives a conclusive result.
For the series $\sum_{n=1}^{\infty} \frac{2^n + (-1)^n}{3^n + n^2}$ , which of the following represents the most appropriate approach using comparison tests?
Answer: Limit comparison with $\sum \left(\frac{2}{3}\right)^n$ since the exponential terms dominate both numerator and denominator For large n, $2^n$ dominates $(-1)^n$ in the numerator and $3^n$ dominates $n^2$ in the denominator. The limit comparison test with $\sum \left(\frac{2}{3}\right)^n$ gives $\lim_{n \to \infty} \frac{\frac{2^n + (-1)^n}{3^n + n^2}}{\left(\frac{2}{3}\right)^n} = \lim_{n \to \infty} \frac{2^n + (-1)^n}{2^n \cdot \frac{3^n + n^2}{3^n}} = 1$ . Since $\sum \left(\frac{2}{3}\right)^n$ converges, the original series converges. Choice A ignores the $(-1)^n$ term incorrectly, C reverses the dominance relationship, and D misapplies the alternating series test to a series that's not alternating.
Let $a_n = \frac{n^2 \sin^2(n)}{n^3 + \sqrt{n}}$ and $b_n = \frac{1}{n}$ . If we apply the direct comparison test to determine the convergence of $\sum_{n=1}^{\infty} a_n$ , what can we conclude?
Answer: The test is inconclusive because we have $0 \leq a_n \leq b_n$ but $\sum b_n$ diverges Since $\sin^2(n) \leq 1$ , we have $a_n \leq \frac{n^2}{n^3 + \sqrt{n}} < \frac{1}{n} = b_n$ for large n. However, the direct comparison test only gives a conclusion when the larger series converges (then the smaller converges) or when the smaller series diverges (then the larger diverges). Here $\sum b_n = \sum \frac{1}{n}$ diverges and $a_n \leq b_n$ , so the test is inconclusive. We cannot determine convergence of $\sum a_n$ from this comparison alone.
The series $\sum_{n=1}^{\infty} \frac{n}{(n+1)3^n}$ is being tested for convergence. Which statement is correct?
Answer: The series converges by Direct Comparison with $\sum \frac{1}{3^n}$ . Let $a_n = \frac{n}{(n+1)3^n}$ . We can use the Direct Comparison Test. For all $n \ge 1$ , the fraction $\frac{n}{n+1}$ is less than 1. Therefore, we can write the inequality: $a_n = \frac{n}{n+1} \cdot \frac{1}{3^n} < 1 \cdot \frac{1}{3^n} = \frac{1}{3^n}$ . Let $b_n = \frac{1}{3^n} = (\frac{1}{3})^n$ . The series $\sum b_n$ is a geometric series with ratio $r = 1/3$ . Since $|r| < 1$ , $\sum b_n$ converges. Because $0 < a_n < b_n$ and $\sum b_n$ converges, the original series $\sum a_n$ must also converge by the Direct Comparison Test.
Consider applying the limit comparison test to $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ where $a_n, b_n > 0$ . If $\lim_{n \to \infty} \frac{a_n}{b_n} = 0$ and $\sum b_n$ converges, what additional information is needed to determine the convergence of $\sum a_n$ ?
Answer: No additional information is needed; $\sum a_n$ must converge by the limit comparison test 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 $\lim_{n \to \infty} \frac{a_n}{b_n} = L$ , then: if $L > 0$ and finite, both series have the same convergence behavior; if $L = 0$ and $\sum b_n$ converges, then $\sum a_n$ also converges; if $L = \infty$ and $\sum b_n$ diverges, then $\sum a_n$ also diverges. In this problem, we have $\lim_{n \to \infty} \frac{a_n}{b_n} = 0$ and $\sum b_n$ 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 $a_n$ approaches zero faster than $b_n$ , so if the "larger" series $\sum b_n$ converges, the "smaller" series $\sum a_n$ 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 $L = 0$ with a convergent comparison series, you immediately know convergence—no additional work required.
The series $\sum_{n=1}^{\infty} \frac{e^{1/n}}{n}$ is analyzed. Which statement is correct?
Answer: The series diverges by Limit Comparison with $\sum \frac{1}{n}$ . Let $a_n = \frac{e^{1/n}}{n}$ . As $n \to \infty$ , $1/n \to 0$ , and $e^{1/n} \to e^0 = 1$ . Therefore, the term $a_n$ behaves like $\frac{1}{n}$ . We use the Limit Comparison Test with the divergent harmonic series $\sum b_n = \sum \frac{1}{n}$ . We calculate the limit of the ratio: $L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{e^{1/n}/n}{1/n} = \lim_{n \to \infty} e^{1/n} = e^0 = 1$ . Since $L=1$ is a finite, positive number and the comparison series $\sum \frac{1}{n}$ diverges, the original series $\sum a_n$ must also diverge.
What is the behavior of the series $\sum_{n=1}^{\infty} \frac{5 + \cos(n\pi)}{n\sqrt{n}}$ ?
Answer: It converges, by Direct Comparison with $\sum \frac{6}{n^{3/2}}$ . The term $\cos(n\pi)$ is equal to $(-1)^n$ . Thus, the numerator $5 + \cos(n\pi)$ alternates between $5-1=4$ and $5+1=6$ . Since the numerator is always positive, all terms of the series are positive, and comparison tests apply. We can bound the general term $a_n$ : $\frac{4}{n\sqrt{n}} \le a_n = \frac{5 + \cos(n\pi)}{n\sqrt{n}} \le \frac{6}{n\sqrt{n}}$ . Let's use the upper bound for a Direct Comparison. We compare our series with $\sum b_n = \sum \frac{6}{n\sqrt{n}} = \sum \frac{6}{n^{3/2}}$ . This is a constant multiple of a p-series with $p=3/2 > 1$ , so it converges. Since $0 \le a_n \le b_n$ and $\sum b_n$ converges, the original series must also converge.
Consider $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}$ where $p > 0$ . Using the integral test as a comparison tool, for which values of p does this series converge?
Answer: $p > 1$ only, because the corresponding improper integral converges precisely when $p > 1$ The integral $\int_2^{\infty} \frac{1}{x(\ln x)^p} dx$ uses substitution $u = \ln x$ , giving $\int_{\ln 2}^{\infty} \frac{1}{u^p} du$ . This converges if and only if $p > 1$ . When $p = 1$ , we get $\int \frac{1}{u} du = \ln u$ , which diverges. When $p < 1$ , the integral also diverges. When $p > 1$ , we get $\frac{u^{1-p}}{1-p}$ which converges. Therefore, the series converges exactly when $p > 1$ .
Which of the following determines the convergence of the series $\sum_{n=1}^{\infty} \sin\left(\frac{1}{n^2}\right)$ ?
Answer: The series converges by the Limit Comparison Test with $\sum \frac{1}{n^2}$ , because $\lim_{x \to 0} \frac{\sin x}{x} = 1$ . For the series $\sum a_n = \sum \sin(1/n^2)$ , we use the Limit Comparison Test. A suitable comparison series is $\sum b_n = \sum \frac{1}{n^2}$ , which is a convergent p-series. We evaluate the limit $\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\sin(1/n^2)}{1/n^2}$ . Let $x = 1/n^2$ ; as $n \to \infty$ , $x \to 0$ . The limit becomes $\lim_{x \to 0} \frac{\sin x}{x} = 1$ . Since the limit is a finite, positive number, and $\sum \frac{1}{n^2}$ converges, the original series also converges.

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