AP Calculus BC Practice Quiz: Alternating Series Test for Convergence
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
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A) The series converges by the Alternating Series Test.
B) The series diverges because the limit of the terms is not zero.
C) The series diverges because the terms are not monotonically decreasing.
D) The Alternating Series Test is inconclusive for this series.
Correct Answer: A
This is an alternating series with \(b_n = \frac{1}{n}\). We check the two conditions for the Alternating Series Test. First, the limit as n approaches infinity of \(b_n\) is \(\lim_{n\to\infty} \frac{1}{n} = 0\). Second, the terms are decreasing, since \(b_{n+1} = \frac{1}{n+1} < \frac{1}{n} = b_n\) for all \(n \ge 1\). Since both conditions are met, the series converges by the Alternating Series Test.
A) The series converges because the terms are alternating.
B) The series converges by the Alternating Series Test.
C) The series diverges because the limit of the terms is not zero.
D) The series diverges because the terms are not decreasing.
Correct Answer: C
To apply the Alternating Series Test, we must first check if the limit of the terms (ignoring the alternating sign) is zero. Let \(b_n = \frac{3n}{2n+1}\). Then \(\lim_{n\to\infty} b_n = \lim_{n\to\infty} \frac{3n}{2n+1} = \frac{3}{2}\). Since the limit is not 0, the series fails the second condition of the Alternating Series Test and diverges by the nth-Term Test for Divergence.
A) \(\lim_{n\to\infty} b_n = 0\) only
B) \(b_{n+1} \le b_n\) for all \(n\) only
C) \(\lim_{n\to\infty} b_n = 0\) and \(b_{n+1} \le b_n\) for all \(n\) (or for \(n\) greater than some integer N)
D) \(\lim_{n\to\infty} (-1)^n b_n = 0\) only
Correct Answer: C
The Alternating Series Test has two conditions that must be met for an alternating series \(\sum (-1)^n b_n\) to converge: (1) the limit of the non-alternating part of the term must be zero, \(\lim_{n\to\infty} b_n = 0\), and (2) the terms \(b_n\) must be non-increasing, i.e., \(b_{n+1} \le b_n\) for all \(n\) past a certain point. Both conditions are required.
A) The series converges by the Alternating Series Test.
B) The series diverges because \(\lim_{n\to\infty} \frac{\ln(n)}{n} \neq 0\).
C) The series diverges because the terms \(b_n = \frac{\ln(n)}{n}\) are not decreasing.
D) The Alternating Series Test cannot be applied to this series.
Correct Answer: A
Let \(b_n = \frac{\ln(n)}{n}\). First, we check the limit: \(\lim_{n\to\infty} \frac{\ln(n)}{n} = 0\) by L'Hôpital's Rule. Second, we check if the terms are decreasing. Let \(f(x) = \frac{\ln(x)}{x}\). The derivative is \(f'(x) = \frac{1 - \ln(x)}{x^2}\), which is negative for \(x > e \approx 2.718\). Thus, the terms \(b_n\) are decreasing for \(n \ge 3\). Since both conditions of the Alternating Series Test are met, the series converges.
A) \(\sum_{n=1}^{\infty} \frac{(-1)^n n^2}{n^2+1}\)
B) \(\sum_{n=1}^{\infty} \frac{(-1)^n (n+1)!}{n!}\)
C) \(\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}+1}\)
D) \(\sum_{n=1}^{\infty} \frac{\cos(n\pi)}{2}\)
Correct Answer: C
We test each series. For A, \(\lim_{n\to\infty} \frac{n^2}{n^2+1} = 1 \neq 0\), so it diverges. For B, \(\frac{(n+1)!}{n!} = n+1\), and \(\lim_{n\to\infty} (n+1) = \infty\), so it diverges. For D, \(\cos(n\pi) = (-1)^n\), so the series is \(\sum \frac{(-1)^n}{2}\). The limit of the terms \(b_n = 1/2\) is not zero, so it diverges. For C, let \(b_n = \frac{1}{\sqrt{n}+1}\). \(\lim_{n\to\infty} b_n = 0\) and the terms are decreasing since the denominator is increasing. Thus, series C converges by the Alternating Series Test.
A) A convergent series.
B) A divergent p-series.
C) A divergent series because \(\lim_{n\to\infty} \frac{\cos(n\pi)}{n^{1/4}} \neq 0\).
D) A series for which the Alternating Series Test is not applicable.
Correct Answer: A
The term \(\cos(n\pi)\) alternates between -1 and 1, equivalent to \((-1)^n\). The series can be rewritten as \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{1/4}}\). This is an alternating series with \(b_n = \frac{1}{n^{1/4}}\). We check the two conditions for the Alternating Series Test. First, \(\lim_{n\to\infty} \frac{1}{n^{1/4}} = 0\). Second, since \(n^{1/4}\) is an increasing function, \(\frac{1}{(n+1)^{1/4}} < \frac{1}{n^{1/4}}\), so the terms are decreasing. Both conditions are met, so the series converges.
A) The series diverges because \(\lim_{n\to\infty} b_n = \infty\).
B) The series diverges because the terms \(b_n\) are not always decreasing.
C) The series converges because \(\lim_{n\to\infty} b_n = 0\) and the terms \(b_n\) are eventually decreasing.
D) The series converges because \(\sum b_n\) converges.
Correct Answer: C
We apply the Alternating Series Test. First, we must evaluate \(\lim_{n\to\infty} b_n = \lim_{n\to\infty} \frac{n^3}{e^n}\). Since the exponential function \(e^n\) grows faster than any polynomial, this limit is 0. (This can be confirmed by applying L'Hôpital's Rule three times). Second, we must determine if the terms are eventually decreasing. Let \(f(x) = \frac{x^3}{e^x}\). Then \(f'(x) = \frac{3x^2e^x - x^3e^x}{(e^x)^2} = \frac{x^2(3-x)}{e^x}\). For \(x > 3\), \(f'(x) < 0\), which means the terms \(b_n\) are decreasing for \(n \ge 4\). Since the terms are eventually decreasing and their limit is 0, the series converges by the Alternating Series Test.