AP Calculus BC Practice Quiz: Comparison Tests for Convergence
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 9 questions to check your progress.
Question 1 of 9
All Questions (9)
A) \(\sum_{n=1}^{\infty} \frac{1}{n^3}\)
B) \(\sum_{n=1}^{\infty} \frac{1}{n}\)
C) \(\sum_{n=1}^{\infty} \frac{1}{n^3-1}\)
D) \(\sum_{n=1}^{\infty} n^3\)
Correct Answer: A
To use the Direct Comparison Test to prove convergence, we need to find a larger, convergent series. For all \(n \ge 1\), we have \(n^3+1 > n^3\), which implies \(\frac{1}{n^3+1} < \frac{1}{n^3}\). The series \(\sum_{n=1}^{\infty} \frac{1}{n^3}\) is a convergent p-series because \(p=3 > 1\). Since the terms of the original series are smaller than the terms of a convergent series, the original series must also converge.
A) The limit is 0, so the series converges.
B) The limit is \(\frac{1}{2}\), so the series diverges.
C) The limit is \(\frac{1}{2}\), so the series converges.
D) The limit is \(\infty\), so the test is inconclusive.
Correct Answer: B
Let \(a_n = \frac{n}{2n^2-1}\) and \(b_n = \frac{1}{n}\). We compute the limit: \(\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n/(2n^2-1)}{1/n} = \lim_{n \to \infty} \frac{n^2}{2n^2-1} = \frac{1}{2}\). Since the limit is a finite, positive number (0 < 1/2 < \(\infty\)), both series share the same convergence behavior. The series \(\sum_{n=1}^{\infty} \frac{1}{n}\) is the harmonic series, which diverges. Therefore, the original series also diverges.
A) \(a_n > 0\)
B) \(\lim_{n \to \infty} a_n = 0\)
C) The partial sums are bounded.
D) The terms \(a_n\) are strictly decreasing.
Correct Answer: A
Both the Direct Comparison Test and the Limit Comparison Test require the terms of the series being tested, as well as the terms of the series used for comparison, to be positive. The other conditions are related to convergence but are not the prerequisite for applying these specific tests.
A) The series converges by direct comparison with \(\sum \frac{1}{\sqrt{n}}\).
B) The series diverges by direct comparison with \(\sum \frac{1}{\sqrt{n}}\).
C) The series converges by the Limit Comparison Test with \(\sum \frac{1}{n^2}\).
D) The series diverges by the Limit Comparison Test with \(\sum \frac{1}{\sqrt{n}}\).
Correct Answer: D
We use the Limit Comparison Test with the divergent p-series \(\sum \frac{1}{\sqrt{n}}\) (where \(p=1/2 \le 1\)). Let \(a_n = \frac{1}{\sqrt{n}-1}\) and \(b_n = \frac{1}{\sqrt{n}}\). Then \(\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{1/(\sqrt{n}-1)}{1/\sqrt{n}} = \lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n}-1} = 1\). Since the limit is a finite positive number, and \(\sum \frac{1}{\sqrt{n}}\) diverges, the series \(\sum_{n=2}^{\infty} \frac{1}{\sqrt{n}-1}\) also diverges. Direct comparison is difficult because \(\frac{1}{\sqrt{n}-1} > \frac{1}{\sqrt{n}}\), which is the correct inequality for proving divergence.
A) If \(\sum b_n\) converges, then \(\sum a_n\) converges.
B) If \(\sum b_n\) diverges, then \(\sum a_n\) diverges.
C) The series \(\sum a_n\) must converge, regardless of \(\sum b_n\).
D) The test is inconclusive.
Correct Answer: A
If \(\lim_{n \to \infty} \frac{a_n}{b_n} = 0\), it means that for large \(n\), the terms \(a_n\) are significantly smaller than the terms \(b_n\). Therefore, if the 'larger' series \(\sum b_n\) converges, the 'smaller' series \(\sum a_n\) must also converge. However, if \(\sum b_n\) diverges, we can make no conclusion about \(\sum a_n\).
A) The series converges by direct comparison with \(\sum \frac{3}{n^2}\).
B) The series diverges by direct comparison with \(\sum \frac{1}{n^2}\).
C) The series converges because \(\lim_{n \to \infty} \frac{2+\cos(n)}{n^2} = 0\).
D) The series diverges by the Limit Comparison Test with \(\sum \frac{1}{n}\).
Correct Answer: A
We know that \(-1 \le \cos(n) \le 1\). Therefore, \(2-1 \le 2+\cos(n) \le 2+1\), which simplifies to \(1 \le 2+\cos(n) \le 3\). This gives the inequality \(0 < \frac{2+\cos(n)}{n^2} \le \frac{3}{n^2}\). The series \(\sum \frac{3}{n^2} = 3\sum \frac{1}{n^2}\) is a constant multiple of a convergent p-series (p=2 > 1), so it converges. By the Direct Comparison Test, since the terms of the original series are less than or equal to the terms of a convergent series, the original series must also converge.
A) Limit Comparison Test with \(\sum \frac{1}{n}\), which shows divergence.
B) Limit Comparison Test with \(\sum \frac{1}{n^2}\), which shows convergence.
C) Direct Comparison Test with \(\sum \frac{1}{n}\), which shows divergence.
D) Direct Comparison Test with \(\sum \frac{1}{n^2}\), which shows convergence.
Correct Answer: A
For this series, the Limit Comparison Test is most effective. We compare it with the harmonic series \(\sum \frac{1}{n}\). We use the known limit \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\). Let \(x = 1/n\). As \(n \to \infty\), \(x \to 0\). So, \(\lim_{n \to \infty} \frac{\sin(1/n)}{1/n} = 1\). Since the limit is a finite, positive number (1), and the harmonic series \(\sum \frac{1}{n}\) diverges, the series \(\sum \sin(\frac{1}{n})\) also diverges.
A) The inequality \(\frac{1}{n^2-1} < \frac{1}{n^2}\) is false; the series diverges.
B) The inequality \(\frac{1}{n^2-1} < \frac{1}{n^2}\) is false; the series converges.
C) The inequality \(\frac{1}{n^2-1} > \frac{1}{n^2}\) is true, so the test is inconclusive; the series diverges.
D) The inequality \(\frac{1}{n^2-1} > \frac{1}{n^2}\) is true, so the test is inconclusive; the series converges.
Correct Answer: D
For \(n \ge 2\), we have \(n^2-1 < n^2\), which means \(\frac{1}{n^2-1} > \frac{1}{n^2}\). The Direct Comparison Test requires the terms of the series in question to be *less than* the terms of a known convergent series. Since the inequality is in the wrong direction, the test is inconclusive with \(\sum \frac{1}{n^2}\). However, we can use the Limit Comparison Test with \(\sum \frac{1}{n^2}\). \(\lim_{n \to \infty} \frac{1/(n^2-1)}{1/n^2} = \lim_{n \to \infty} \frac{n^2}{n^2-1} = 1\). Since the limit is 1 and \(\sum \frac{1}{n^2}\) converges, the original series also converges.
A) \(\sum_{n=1}^{\infty} \frac{1}{n+5}\)
B) \(\sum_{n=1}^{\infty} \frac{1}{n^2}\)
C) \(\sum_{n=1}^{\infty} \frac{\ln(n)}{n}\)
D) \(\sum_{n=1}^{\infty} \frac{1}{2^n}\)
Correct Answer: C
To prove divergence with the Direct Comparison Test, we must show that the terms of our series are greater than the terms of a known divergent series. We compare with \(\sum \frac{1}{n}\). For \(n \ge 3\), \(\ln(n) > 1\), which means \(\frac{\ln(n)}{n} > \frac{1}{n}\). Since the terms of \(\sum \frac{\ln(n)}{n}\) are larger than the terms of the divergent harmonic series (for \(n \ge 3\)), the series \(\sum \frac{\ln(n)}{n}\) must also diverge. For option A, \(\frac{1}{n+5} < \frac{1}{n}\), so the test is inconclusive.