# Power Series Practice Problems with Answers and Solutions

> 12 power series practice problems with answers: radius and interval of convergence, endpoint checks, and building series from 1/(1 − x).

Source: https://practicecalculus.com/practice-problems/power-series
Updated: 2026-11-06

$$ \frac{1}{1 - x} = \sum_{n=0}^{\infty}x^n, \quad |x| < 1 $$

## At a glance

- **Problems:** 12, with answers and full solutions
- **Level:** Calculus 2 and AP Calculus BC, Unit 10
- **You need:** The [ratio test](/rules/ratio-test) and the other [convergence tests](/practice-problems/convergence-tests)
- **Main idea:** Use the ratio test for the radius, then test each endpoint on its own
- **Practice with new numbers:** [Power series and radius lesson](/learn/calculus-2/power-series-parametric-polar/power-series-and-radius)

## Key takeaways

- Apply the ratio test to $|a_n(x - c)^n|$ and solve "limit $< 1$" for $x$ to get the radius.
- The ratio test says nothing at the endpoints. Plug each one in and use another test.
- Build new series from $\frac{1}{1 - x}$ by substituting, multiplying by powers of $x$, differentiating or integrating. The radius stays the same.

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:

$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| < 1 \quad\text{for the radius}, \qquad \frac{1}{1 - x} = \sum_{n=0}^{\infty}x^n \quad\text{for building series}$$

For more examples, read the [power series guide](/blog/power-series).

## 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

| Problem | Radius | Interval or answer |
| --- | --- | --- |
| 1 | $1$ | $[-1, 1)$ |
| 2 | $\infty$ | $(-\infty, \infty)$ |
| 3 | $0$ | Only $x = 0$ |
| 4 | $3$ | $(-1, 5)$ |
| 5 | $1$ | $[-1, 1]$ |
| 6 | $2$ | $(-2, 2)$ |
| 7 | $1$ | $(-2, 0)$ |
| 8 | $\frac{1}{2}$ | $[0, 1)$ |
| 9 | $\frac{1}{3}$ | $\sum 3^n x^n$ |
| 10 | $1$ | $\sum(-1)^n x^{2n+1}$ |
| 11 | $1$ | $\sum n x^{n-1}$; the sum is $2$ |
| 12 | $1$ | $\sum\frac{(-1)^{n+1}x^n}{n}$ |

## 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:

$$\frac{1}{1 - 3x} = \sum_{n=0}^{\infty}(3x)^n = \sum_{n=0}^{\infty}3^n x^n, \quad |3x| < 1 \Rightarrow R = \frac{1}{3}$$

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

$$\frac{x}{1 + x^2} = x\sum_{n=0}^{\infty}(-x^2)^n = \sum_{n=0}^{\infty}(-1)^n x^{2n+1}, \quad |x| < 1$$

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

$$\frac{1}{(1 - x)^2} = \sum_{n=1}^{\infty}n x^{n-1}$$

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 $x$:

$$\ln(1 + x) = \sum_{n=0}^{\infty}\frac{(-1)^n x^{n+1}}{n + 1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$$

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](/learn/calculus-2/power-series-parametric-polar/power-series-and-radius). Next, learn to build series for any function in the [Taylor series problems](/practice-problems/taylor-series).

## Frequently asked questions

### What is the difference between the radius and the interval of convergence?

The radius $R$ is how far from the center the series converges. The interval is the actual set of $x$-values, including which endpoints belong. Two series with the same radius can have different intervals, as problems 1 and 5 show.

### Why do you have to check the endpoints separately?

At an endpoint the ratio test limit equals exactly $1$, which gives no information. The series becomes an ordinary number series, and it might converge or diverge.

### Does differentiating a power series change the radius?

No. Differentiating or integrating term by term keeps the same radius. It can change what happens at the endpoints, so check them again.