Power Series — Radius of Convergence & Examples

A power series is an infinite sum built from powers of $(x-c)$:

$$\sum_{n=0}^{\infty} a_n (x-c)^n$$

Here, $c$ is the center and the numbers $a_n$ are constants called coefficients. In most problems, the real question is simple: for which values of $x$ does this series converge?

The answer is organized by the radius of convergence $R$. A power series converges when $|x-c| < R$, diverges when $|x-c| > R$, and needs separate endpoint checks when $|x-c| = R$.

What Radius Of Convergence Means

The radius of convergence is a distance from the center, not a set of $x$-values. If a power series is centered at $c$, then:

• it converges when $|x-c| < R$,
• it diverges when $|x-c| > R$,
• the boundary case $|x-c| = R$ must be tested separately.

For real-variable problems, that distance becomes an interval of convergence. If the center is $c$ and the radius is $R$, the inside part is

$$(c-R,\; c+R),$$

but the endpoints may or may not be included in the final answer.

Why Power Series Matter

Power series matter because they let you treat complicated functions like very long polynomials. Inside the convergence interval, they are often easier to differentiate, integrate, and approximate.

That shortcut comes with a condition: those term-by-term operations are justified inside the interval of convergence, not automatically everywhere.

Power Series Example: Find The Radius And Interval

Consider

$$\sum_{n=0}^{\infty} \frac{(x-2)^n}{3^n}.$$

This is a power series centered at $c=2$. To find the radius of convergence, apply the ratio test to

$$a_n = \frac{(x-2)^n}{3^n}.$$

Compute

$$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(x-2)^{n+1}}{3^{n+1}} \cdot \frac{3^n}{(x-2)^n}\right| = \frac{|x-2|}{3}.$$

The ratio test gives convergence when

$$\frac{|x-2|}{3} < 1,$$

so

$$|x-2| < 3.$$

So the radius of convergence is

$$R = 3.$$

That gives the interior interval $(-1,5)$. Now test the endpoints one at a time.

At $x=5$, the series becomes

$$\sum_{n=0}^{\infty} 1,$$

which diverges.

At $x=-1$, the series becomes

$$\sum_{n=0}^{\infty} (-1)^n,$$

which also diverges because its terms alternate between $1$ and $-1$ instead of approaching $0$.

So the final interval of convergence is

$$(-1,5).$$

This is the full workflow in one example: identify the center, find $R$, write the inside interval, and then test both endpoints separately.

Common Mistakes With Radius Of Convergence

Mixing Up Radius And Interval

The radius is a number such as $R=3$. The interval is the set of real $x$-values, such as $(-1,5)$. They are related, but they are not the same object.

Forgetting The Center $c$

In $\sum a_n (x-c)^n$, the center is $c$, not always $0$. If the series uses $(x-2)^n$, the distance test is based on $|x-2|$, not $|x|$.

Skipping The Endpoint Tests

The ratio test and root test usually tell you what happens for the interior and exterior, but they often say nothing at the endpoints. You still have to check them one at a time.

Assuming Both Endpoints Behave The Same Way

Even if the radius is the same on both sides, one endpoint may converge while the other diverges. Endpoint behavior depends on the series you get after substitution.

When Power Series Are Used

Power series appear in calculus, differential equations, and approximation. They are useful when a function is hard to handle directly but easier to study near one point through its series expansion.

Taylor and Maclaurin series are important examples. They are power series designed to represent a function locally, when the needed conditions are met.

Try A Similar Power Series

Try your own version with

$$\sum_{n=0}^{\infty} \frac{(x+1)^n}{2^n}.$$

Find the center, solve for the radius, and then test the endpoints. If you want one more nearby case after that, explore a Taylor series and notice how the same convergence ideas appear again.