AP Calculus BC Practice Quiz: Harmonic Series and $p$-Series
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 7 questions to check your progress.
Question 1 of 7
All Questions (7)
A) $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$
B) $\sum_{n=1}^{\infty} \frac{1}{n}$
C) $\sum_{n=1}^{\infty} \frac{1}{n^{0.99}}$
D) $\sum_{n=1}^{\infty} \frac{1}{n^2}$
Correct Answer: D
A series of the form $\sum \frac{1}{n^p}$ is a p-series, which converges if and only if $p > 1$. In option A, $p=0.5$. In option B, $p=1$ (the harmonic series). In option C, $p=0.99$. In option D, $p=2$. Only option D has an exponent $p > 1$, so it is the only series that converges.
A) $\sum_{n=1}^{\infty} \frac{1}{n^{1.01}}$
B) $\sum_{n=1}^{\infty} \frac{1}{n \sqrt{n}}$
C) $\sum_{n=1}^{\infty} \frac{1}{n}$
D) $\sum_{n=1}^{\infty} \frac{1}{n^3}$
Correct Answer: C
A p-series of the form $\sum \frac{1}{n^p}$ diverges if $p \le 1$. Option A has $p=1.01 > 1$, so it converges. Option B can be written as $\sum \frac{1}{n^{1.5}}$, so $p=1.5 > 1$ and it converges. Option D has $p=3 > 1$, so it converges. Option C is the harmonic series, which is a p-series with $p=1$, and therefore it diverges.
A) The series converges because it is a p-series with $p = 5/3$.
B) The series diverges because it is a p-series with $p = 5/3$.
C) The series converges because it is a p-series with $p = 3/5$.
D) The series diverges because it is a p-series with $p = 3/5$.
Correct Answer: A
The general term of the series can be rewritten using exponent rules: $\frac{1}{\sqrt[3]{n^5}} = \frac{1}{n^{5/3}}$. This shows the series is a p-series with $p = 5/3$. Since $p = 5/3 \approx 1.67$, which is greater than 1, the p-series converges.
A) $k > 1$
B) $k > 1/2$
C) $k < 1$
D) $k < 1/2$
Correct Answer: B
This is a p-series with $p = 2k$. For a p-series to converge, the exponent $p$ must be greater than 1. Therefore, we must have the inequality $2k > 1$. Solving for $k$ gives $k > 1/2$.
A) The series converges to 0.
B) The series converges to a value greater than 1.
C) The series diverges.
D) The series is a geometric series and converges.
Correct Answer: C
The harmonic series is a specific type of p-series, $\sum \frac{1}{n^p}$, where $p=1$. According to the p-series test, a series diverges if $p \le 1$. Since $p=1$ for the harmonic series, it diverges.
A) Both $\sum_{n=1}^{\infty} \frac{1}{n}$ and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converge.
B) Both $\sum_{n=1}^{\infty} \frac{1}{n}$ and $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ diverge.
C) $\sum_{n=1}^{\infty} \frac{1}{n}$ converges, but $\sum_{n=1}^{\infty} \frac{1}{n^2}$ diverges.
D) $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, but $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ also converges.
Correct Answer: B
We use the p-series test, which states that $\sum \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \le 1$. 1. $\sum \frac{1}{n}$ is a p-series with $p=1$, so it diverges. 2. $\sum \frac{1}{n^2}$ is a p-series with $p=2$, so it converges. 3. $\sum \frac{1}{\sqrt{n}}$ is a p-series with $p=1/2$, so it diverges. Based on this, statement B is the only true statement, as both series listed have $p \le 1$.
A) It diverges because the harmonic series $\sum \frac{1}{n}$ diverges.
B) It converges.
C) It diverges because the terms do not approach zero.
D) It is a p-series with $p=1$ and therefore diverges.
Correct Answer: B
The alternating harmonic series is not a p-series because of the alternating sign. While the series of its absolute values, the harmonic series $\sum \frac{1}{n}$, diverges, the alternating series itself converges. This can be shown by the Alternating Series Test, as the terms $\frac{1}{n}$ are positive, decreasing, and have a limit of 0 as $n \to \infty$.