Practice Problems – Infinite Sequences & Series – AP Calculus BC

Below are 20 practice problems covering all topics from Module 06: Infinite Sequences & Series (BC). Topics include: sequences, geometric series, integral test, comparison tests, ratio/root tests, alternating series, absolute/conditional convergence. Each problem includes a hidden solution — click the button to reveal the step-by-step answer.

Problem 01 Sequence

Find $\displaystyle \lim_{n \to \infty} \frac{2n+1}{n+3}$.

Solution:
$\lim_{n \to \infty} \frac{2n+1}{n+3} = \lim_{n \to \infty} \frac{2 + 1/n}{1 + 3/n} = 2$

Problem 02 Geometric Series

Find the sum of $\sum_{n=0}^\infty \frac{1}{3^n}$.

Solution:
$a=1$, $r=1/3$, $|r|<1$, sum $= \frac{1}{1-1/3} = \frac{1}{2/3} = \frac{3}{2}$

Problem 03 Integral Test

Use the Integral Test to determine convergence of $\sum_{n=1}^\infty \frac{1}{n^3}$.

Solution:
$\int_1^\infty x^{-3} dx = \lim_{b\to\infty} [-\frac{1}{2x^2}]_1^b = \lim_{b\to\infty} (-\frac{1}{2b^2} + \frac{1}{2}) = \frac{1}{2}$ (converges). So the series converges.

Problem 04 p-Series

Determine if $\sum_{n=1}^\infty \frac{1}{\sqrt{n}}$ converges or diverges.

Solution:
$p = 1/2 < 1$, so the p-series diverges.

Problem 05 Direct Comparison

Does $\sum_{n=1}^\infty \frac{1}{n^2+1}$ converge or diverge?

Solution:
$\frac{1}{n^2+1} \le \frac{1}{n^2}$ and $\sum 1/n^2$ converges, so the series converges.

Problem 06 Limit Comparison

Determine convergence of $\sum_{n=1}^\infty \frac{2n+1}{n^2+3}$.

Solution:
Compare to $b_n = 1/n$. $\lim \frac{(2n+1)/(n^2+3)}{1/n} = \lim \frac{2n^2+n}{n^2+3} = 2$. Since $\sum 1/n$ diverges, the series diverges.

Problem 07 Ratio Test

Test $\sum_{n=1}^\infty \frac{2^n}{n!}$ for convergence.

Solution:
$\lim \frac{2^{n+1}/(n+1)!}{2^n/n!} = \lim \frac{2}{n+1} = 0 < 1$ → converges absolutely.

Problem 08 Root Test

Test $\sum_{n=1}^\infty \left(\frac{3n}{4n+1}\right)^n$ for convergence.

Solution:
$|a_n|^{1/n} = \frac{3n}{4n+1} \to \frac{3}{4} < 1$ → converges.

Problem 09 Alternating Series

Does $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}$ converge absolutely, conditionally, or diverge?

Solution:
$\sum 1/n^2$ converges → absolutely convergent.

Problem 10 Alternating Series Error

For $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}$, find an upper bound for the error after 5 terms.

Solution:
Error $|R_5| \le b_6 = 1/6 \approx 0.1667$.

Problem 11 Absolute/Conditional

Does $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{\sqrt{n}}$ converge absolutely, conditionally, or diverge?

Solution:
$\sum 1/\sqrt{n}$ diverges (p=1/2). Alternating series converges (decreasing to 0). So conditionally convergent.

Problem 12 Geometric Series

Find the sum of $\sum_{n=1}^\infty 2\left(\frac{1}{3}\right)^n$.

Solution:
$a = 2/3$, $r=1/3$, sum $= \frac{2/3}{1-1/3} = \frac{2/3}{2/3} = 1$

Problem 13 Telescoping Series

Find the sum of $\sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right)$.

Solution:
$S_N = 1 - 1/(N+1) \to 1$ as $N\to\infty$. Sum = 1.

Problem 14 Ratio Test

What does the Ratio Test say about $\sum_{n=1}^\infty \frac{1}{n}$?

Solution:
$\lim \frac{1/(n+1)}{1/n} = \lim \frac{n}{n+1} = 1$ → inconclusive.

Problem 15 Sequence Limit

Find $\lim_{n \to \infty} \frac{\ln n}{n}$.

Solution:
L'Hôpital: $\lim_{x \to \infty} \frac{1/x}{1} = 0$.

Problem 16 Limit Comparison

Determine convergence of $\sum_{n=1}^\infty \frac{3n^2+2}{n^4+1}$.

Solution:
Compare to $1/n^2$: $\lim \frac{(3n^2+2)/(n^4+1)}{1/n^2} = \lim \frac{3n^4+2n^2}{n^4+1} = 3$. Since $\sum 1/n^2$ converges, the series converges.

Problem 17 Alternating Series

Does $\sum_{n=1}^\infty \frac{(-1)^{n-1} n}{n+1}$ converge?

Solution:
$\lim b_n = \lim \frac{n}{n+1} = 1 \neq 0$, so the series diverges (nth term test).

Problem 18 Geometric Series

Does $\sum_{n=0}^\infty 3 \cdot (2)^n$ converge or diverge?

Solution:
$r=2 > 1$, so the geometric series diverges.

Problem 19 Root Test

Test $\sum_{n=1}^\infty \left(\frac{1}{n}\right)^n$ for convergence.

Solution:
$|a_n|^{1/n} = 1/n \to 0 < 1$ → converges.

Problem 20 Conditional Convergence

Give an example of a conditionally convergent series.

Solution:
The alternating harmonic series $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}$ is conditionally convergent.

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