Power Series Practice Problems with Answers and Solutions

These power series practice problems cover radius and interval of convergence, endpoint checks, and building new series from the geometric series. Try each one first. The answers come first, then full solutions.

The two tools:

For more examples, read the power series guide.

Problems

Radius and interval of convergence

1. \displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n}
2. \displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}
3. \displaystyle\sum_{n=0}^{\infty}n!\,x^n
4. \displaystyle\sum_{n=0}^{\infty}\frac{(x - 2)^n}{3^n}
5. \displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n^2}
6. \displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{4^n}
7. \displaystyle\sum_{n=1}^{\infty}n(x + 1)^n
8. \displaystyle\sum_{n=1}^{\infty}\frac{(2x - 1)^n}{n}

Building series

9. Find a power series for \dfrac{1}{1 - 3x} and its radius.
10. Find a power series for \dfrac{x}{1 + x^2} and its radius.
11. Differentiate \frac{1}{1 - x} to get a series for \frac{1}{(1 - x)^2}. Use it to find \displaystyle\sum_{n=1}^{\infty}\frac{n}{2^n}.
12. Integrate a series for \frac{1}{1 + x} to get a series for \ln(1 + x).

Answers

Step-by-step solutions

1. Ratio: \left|\frac{x^{n+1}}{n + 1}\cdot\frac{n}{x^n}\right| = \frac{n}{n + 1}|x| \to |x|. So |x| < 1 and R = 1. At x = 1: \sum\frac{1}{n} diverges. At x = -1: \sum\frac{(-1)^n}{n} converges by the alternating series test. Interval [-1, 1).

2. Ratio: \frac{|x|}{n + 1} \to 0 < 1 for every x. So R = \infty. This is the series for e^x.

3. Ratio: (n + 1)|x| \to \infty for every x \neq 0. The series only converges at its center.

4. This is geometric with ratio \frac{x - 2}{3}. It converges when |x - 2| < 3, which is -1 < x < 5. At both endpoints the terms are (\pm 1)^n, which do not go to 0, so both diverge.

5. Ratio: \frac{n^2}{(n + 1)^2}|x| \to |x|, so R = 1. At x = \pm 1, \sum\frac{(\pm 1)^n}{n^2} converges absolutely, since \sum\frac{1}{n^2} converges. Both endpoints are in.

6. This is geometric with ratio -\frac{x^2}{4}. It converges when \frac{x^2}{4} < 1, which is |x| < 2. At x = \pm 2 the terms are (-1)^n, which diverge.

7. Ratio: \frac{n + 1}{n}|x + 1| \to |x + 1|, so the series converges for |x + 1| < 1, which is -2 < x < 0. At both endpoints the terms n(\pm 1)^n grow, so both diverge.

8. Ratio: \frac{n}{n + 1}|2x - 1| \to |2x - 1|. Solve |2x - 1| < 1: 0 < x < 1, with center \frac{1}{2} and R = \frac{1}{2}. At x = 1: \sum\frac{1}{n} diverges. At x = 0: \sum\frac{(-1)^n}{n} converges. Interval [0, 1).

9. Replace x with 3x in the geometric series:

10. Replace x with -x^2, then multiply by x:

11. Differentiate both sides of \frac{1}{1 - x} = \sum x^n:

Multiply by x: \sum n x^n = \frac{x}{(1 - x)^2}. At x = \frac{1}{2}, this is \frac{1/2}{1/4} = 2.

12. \frac{1}{1 + x} = \sum(-1)^n x^n. Integrate from 0 to x:

Shifting the index gives the form in the table. The radius stays 1.

Common mistakes on these problems

• Skipping the endpoints. Problems 1, 5 and 8 all have R = 1 or R = \frac{1}{2}, but different endpoint behavior.
• Forgetting the center. In problem 4 the center is 2, so the interval is 2 \pm 3, not \pm 3.
• Wrong radius from 2x. In problem 8, |2x - 1| < 1 means |x - \frac{1}{2}| < \frac{1}{2}, so R = \frac{1}{2}, not 1.
• Leaving x in the limit. In the ratio test, |x| is a constant. Take the limit in n, then solve for x.

Keep practicing

For unlimited series with instant feedback, try the power series and radius lesson. Next, learn to build series for any function in the Taylor series problems.