Power Series: Radius and Interval of Convergence

Q: What is a power series?

A series of the form c₀ + c₁(x − a) + c₂(x − a)² + ⋯, with a variable x. Its center is a.

Q: What is the radius of convergence?

The number R such that the series converges when |x − a| R. It can be 0 or ∞.

Q: Why do I have to test the endpoints separately?

At the endpoints the ratio test gives L = 1 and decides nothing. Each endpoint gives a series of numbers that must be tested on its own.

Q: Can a power series converge at one endpoint but not the other?

Yes. The sum of xⁿ/n converges at x = −1, where it alternates, and diverges at x = 1, where it is the harmonic series.

Quick answer

A power series, the sum of cₙ(x − a)ⁿ, is a polynomial that never ends. For each x it is a series of numbers, and the x-values where it converges form an interval centered at a. It converges absolutely when |x − a| < R, diverges when |x − a| > R, and at the two endpoints anything can happen. The ratio test finds the radius R; each endpoint then needs its own test, often a p-series or the alternating series test.

What you'll learn

Recognize a power series and its center
Find the radius of convergence with the ratio test
Test the endpoints to find the interval of convergence
Identify series that converge everywhere or only at the center

Series with a variable

A power series centered at $a$ is

\sum_{n=0}^{\infty} c_n (x - a)^n = c_0 + c_1(x - a) + c_2(x - a)^2 + \cdots

a polynomial that never ends. For each value of $x$ it is a series of numbers, which converges or not. The geometric series is the model: $\sum_{n=0}^{\infty} x^n = \tfrac{1}{1 - x}$ exactly when $\lvert x \rvert < 1$ .

Where a power series converges

For every power series centered at $a$ there is a radius of convergence $R$ , with $0 \le R \le \infty$ , such that the series converges absolutely when $\lvert x - a \rvert < R$ and diverges when $\lvert x - a \rvert > R$ .

The $x$ -values where the series converges form the interval of convergence: from $a - R$ to $a + R$ , with each endpoint in or out.

Why the region is an interval

The ratio test explains the shape. For a power series, the ratio of consecutive terms typically approaches $L\lvert x - a \rvert$ for some number $L$ . The test demands $L\lvert x - a \rvert < 1$ , which says $\lvert x - a \rvert < \tfrac{1}{L}$ : a condition on the distance from the center, the same in both directions. A power series converges within a fixed distance of its center, because each term is the one before times something proportional to $x - a$ . At distance exactly $R$ the ratio test gives $1$ , so the endpoints are left open to further testing.

Testing the endpoints

At the endpoints the ratio test gives $L = 1$ . Substitute each endpoint and test the resulting series of numbers with anything that fits.

For $\sum \tfrac{x^n}{n}$ : at $x = 1$ it is the harmonic series, which diverges. At $x = -1$ it is $\sum \tfrac{(-1)^n}{n}$ , which converges by the alternating series test. So the interval of convergence is $[-1, 1)$ .

The interval of convergence of the series of xⁿ/n

Worked examples

Common mistakes

Practice problems

Find the interval of convergence of $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{3^n}$ .

Answer

$(-3, 3)$

Full solution

It is geometric with ratio $\tfrac{x}{3}$ , which converges exactly when $\lvert x \rvert < 3$ . At $x = \pm 3$ the terms are $1$ or $\pm 1$ , which do not approach $0$ .
Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} n x^n$ .

Answer

$(-1, 1)$

Full solution

The ratio is $\tfrac{n + 1}{n}\lvert x \rvert \to \lvert x \rvert$ , so $R = 1$ . At $x = \pm 1$ the terms $n$ or $\pm n$ do not approach $0$ .
Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n^2}$ .

Answer

$[-1, 1]$

Full solution

$R = 1$ by the ratio test. At $x = 1$ , $\sum \tfrac{1}{n^2}$ converges; at $x = -1$ , $\sum \tfrac{(-1)^n}{n^2}$ converges absolutely.
Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(x - 2)^n}{n}$ .

Answer

$[1, 3)$

Full solution

$R = 1$ , centered at $2$ . At $x = 3$ it is the harmonic series; at $x = 1$ it is the alternating harmonic series, which converges.
Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(2x)^n}{n}$ .

Answer

$\left[-\tfrac{1}{2}, \tfrac{1}{2}\right)$

Full solution

The ratio approaches $2\lvert x \rvert$ , so $\lvert x \rvert < \tfrac{1}{2}$ . At $x = \tfrac{1}{2}$ it is the harmonic series; at $x = -\tfrac{1}{2}$ it is the alternating harmonic series.
Find the interval of convergence of $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n + 1}$ .

Answer

$(-1, 1]$

Full solution

$R = 1$ . At $x = 1$ it is $\sum \tfrac{(-1)^n}{n + 1}$ , which converges; at $x = -1$ it is $\sum \tfrac{1}{n + 1}$ , which diverges.
Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n\,4^n}$ .

Answer

$[-4, 4)$

Full solution

The ratio approaches $\tfrac{\lvert x \rvert}{4}$ , so $R = 4$ . At $x = 4$ it is $\sum \tfrac{1}{n}$ ; at $x = -4$ it is $\sum \tfrac{(-1)^n}{n}$ .
Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(x + 1)^n}{\sqrt{n}}$ .

Answer

$[-2, 0)$

Full solution

$R = 1$ , centered at $-1$ . At $x = 0$ it is $\sum \tfrac{1}{\sqrt{n}}$ , which diverges; at $x = -2$ it is $\sum \tfrac{(-1)^n}{\sqrt{n}}$ , which converges.
Find where $\displaystyle\sum_{n=0}^{\infty} n!\,(x - 5)^n$ converges.

Answer

Only at $x = 5$

Full solution

The ratio is $(n + 1)\lvert x - 5 \rvert$ , which grows without bound unless $x = 5$ . So $R = 0$ .
A student finds $R = 1$ for $\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n}$ and gives the interval of convergence as $(-1, 1)$ . What went wrong?

Hint

What series do you get at $x = -1$ ?

Answer

The endpoints were not tested. The series converges at $x = -1$ , so the interval is $[-1, 1)$ .

Full solution

At $x = -1$ the series is $\sum \tfrac{(-1)^n}{n}$ , which converges by the alternating series test. At $x = 1$ it is the harmonic series, which diverges.

Frequently asked questions

What is a power series?

A series of the form c₀ + c₁(x − a) + c₂(x − a)² + ⋯, with a variable x. Its center is a.

What is the radius of convergence?

The number R such that the series converges when |x − a| < R and diverges when |x − a| > R. It can be 0 or ∞.

How do I find the radius of convergence?

Apply the ratio test to the absolute values of the terms, with x in them, and solve L < 1 for |x − a|.

Why do I have to test the endpoints separately?