# Power Series: Radius and Interval of Convergence, With Examples

> Find the radius and interval of convergence of a power series with the ratio test, check both endpoints correctly, and follow five fully worked examples.

Source: https://practicecalculus.com/blog/power-series
Updated: 2026-09-20

## Key takeaways

- A power series $\sum c_n (x-a)^n$ always converges on an interval centered at $a$, and its half-width is the radius of convergence $R$.
- The ratio test gives the radius: set $\lim_{n\to\infty} \left|\frac{c_{n+1}}{c_n}\right| |x-a| < 1$ and solve for $|x-a|$.
- The ratio test always says nothing at the endpoints, so plug each endpoint in and test the resulting number series separately.
- The radius can be $0$, a positive number, or $\infty$, and the two endpoints can behave differently.

You know how to decide whether a series of numbers converges. A power series adds a twist: it has an $x$ in every term, so it converges for some values of $x$ and diverges for others. Your job is to find exactly which ones, and the ratio test does almost all of the work.

This guide shows you how to find the radius and interval of convergence step by step, why the endpoints need extra care, and five worked examples that cover every case you will meet on an exam.

## What is a power series?

A **power series centered at $a$** is an infinite sum of powers of $(x - a)$, each with its own coefficient:

$$\sum_{n=0}^{\infty} c_n (x-a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + \cdots$$

Think of it as a polynomial that never ends. For each number you plug in for $x$, you get an ordinary series of numbers, which either converges or diverges.

The most famous example is the geometric series

$$\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots = \frac{1}{1-x}, \quad |x| < 1$$

It works perfectly inside $(-1, 1)$ and fails everywhere else. That pattern, a clean interval around the center, is true of every power series.

```graph
title: 'Partial sums of the geometric power series 1 + x + x² + ... compared with 1/(1 − x)'
caption: 'Inside $(-1, 1)$ the partial sums hug $\frac{1}{1-x}$. Outside that interval they peel away, because the series diverges there.'
x: [-1.8, 1.4]
y: [-2, 6]
functions:
    - expr: 1/(1-x)
      label: y = 1/(1 − x)
    - expr: 1 + x + x^2 + x^3
      label: degree 3 partial sum
      dashed: true
    - expr: 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8
      label: degree 8 partial sum
      dashed: true
vertical:
    - {x: -1, label: 'x = −1'}
    - {x: 1, label: 'x = 1'}
```

## Radius of convergence: the three possibilities

For every power series centered at $a$, exactly one of these is true:

1. It converges **only at $x = a$**. The radius is $R = 0$.
2. It converges **for every $x$**. The radius is $R = \infty$.
3. There is a number $R > 0$ so that it converges when $|x - a| < R$ and diverges when $|x - a| > R$.

The number $R$ is the **radius of convergence**. The **interval of convergence** is the set of all $x$ where the series converges. In case 3 that is the interval from $a - R$ to $a + R$, plus whichever endpoints happen to work.

## How to find the radius of convergence with the ratio test

The ratio test says a series $\sum b_n$ converges absolutely if $\lim_{n\to\infty} \left|\frac{b_{n+1}}{b_n}\right| < 1$ and diverges if that limit is greater than 1. Apply it to the terms $b_n = c_n (x-a)^n$:

$$L = \lim_{n\to\infty} \left| \frac{c_{n+1}(x-a)^{n+1}}{c_n (x-a)^n} \right| = |x-a| \lim_{n\to\infty} \left|\frac{c_{n+1}}{c_n}\right|$$

Then set $L < 1$ and solve for $|x - a|$. The recipe:

1. Write the ratio $\left|\frac{b_{n+1}}{b_n}\right|$ and simplify it. The powers of $(x - a)$ cancel down to a single $|x - a|$.
2. Take the limit as $n \to \infty$. The $x$ stays put because it does not depend on $n$.
3. Solve $L < 1$. You get $|x - a| < R$.
4. **Check both endpoints** $x = a - R$ and $x = a + R$ by plugging them in.

## Why the endpoints need a separate check

At $x = a \pm R$ the ratio test limit equals exactly 1, and a limit of 1 tells you nothing. The series might converge or diverge, and the two endpoints do not have to agree. So at each endpoint you substitute the number into the series and use a different test: the p-series test, the alternating series test, the nth-term test or a comparison. If you need a refresher on those, see our guide to [convergence tests](/blog/convergence-tests).

> **Tip:** never skip the endpoints. An answer of $(-1, 1)$ when the truth is $[-1, 1)$ is marked wrong, even though the radius is right.

## Worked example 1: endpoints that disagree

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n}$.

**Step 1. Set up the ratio.**

$$\left|\frac{x^{n+1}/(n+1)}{x^n/n}\right| = |x| \cdot \frac{n}{n+1}$$

**Step 2. Take the limit.** Since $\frac{n}{n+1} \to 1$, we get $L = |x|$.

**Step 3. Solve $L < 1$.** The series converges for $|x| < 1$, so $R = 1$.

**Step 4. Check the endpoints.**

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

The interval of convergence is $[-1, 1)$.

## Worked example 2: a series not centered at 0

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(x-3)^n}{n^2\, 2^n}$.

**Step 1. Set up the ratio.**

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

**Step 2. Take the limit.** The fraction $\frac{n^2}{(n+1)^2}$ tends to 1, so $L = \frac{|x-3|}{2}$.

**Step 3. Solve $L < 1$.** We need $|x - 3| < 2$, so $R = 2$ and the open interval is $1 < x < 5$.

**Step 4. Check the endpoints.**

- At $x = 5$: $(x-3)^n = 2^n$, and the series becomes $\sum \frac{2^n}{n^2\, 2^n} = \sum \frac{1}{n^2}$, a p-series with $p = 2$. It **converges**.
- At $x = 1$: $(x-3)^n = (-2)^n$, and the series becomes $\sum \frac{(-1)^n}{n^2}$. It converges absolutely, since $\sum \frac{1}{n^2}$ converges.

The interval of convergence is $[1, 5]$. Notice that the center is 3 and the interval stretches 2 units each way.

## Worked example 3: both endpoints diverge

Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{n\,(x+1)^n}{4^n}$.

**Step 1. Set up the ratio.**

$$\left|\frac{(n+1)(x+1)^{n+1}}{4^{n+1}} \cdot \frac{4^n}{n\,(x+1)^n}\right| = \frac{|x+1|}{4} \cdot \frac{n+1}{n}$$

**Step 2. Take the limit.** $L = \frac{|x+1|}{4}$.

**Step 3. Solve $L < 1$.** $|x + 1| < 4$, so $R = 4$ and the center is $a = -1$. That gives $-5 < x < 3$.

**Step 4. Check the endpoints.**

- At $x = 3$: the series is $\sum \frac{n \cdot 4^n}{4^n} = \sum n$. The terms grow, so it **diverges** by the nth-term test.
- At $x = -5$: the series is $\sum n(-1)^n$. The terms do not approach 0, so it also **diverges**.

The interval of convergence is $(-5, 3)$.

> **Watch the center.** In $(x + 1)^n$ the center is $a = -1$, not $1$. Rewrite it as $(x - (-1))^n$ if that helps.

## Worked example 4: radius infinity and radius zero

Factorials push the radius to one of the two extremes.

**Series A:** $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!}$.

$$\left|\frac{x^{n+1}}{(n+1)!} \cdot \frac{n!}{x^n}\right| = \frac{|x|}{n+1} \to 0$$

The limit is 0 for every $x$, and $0 < 1$ always. So $R = \infty$ and the interval is $(-\infty, \infty)$. There are no endpoints to check. (This series equals $e^x$, as you will see in our guide to [Taylor series](/blog/taylor-series).)

**Series B:** $\displaystyle\sum_{n=0}^{\infty} n!\, x^n$.

$$\left|\frac{(n+1)!\, x^{n+1}}{n!\, x^n}\right| = (n+1)|x|$$

If $x \ne 0$ this grows without bound, so the series diverges. It converges only at the center. So $R = 0$ and the "interval" is the single point $\{0\}$.

## Worked example 5: a coefficient on x

Find the radius of convergence of $\displaystyle\sum_{n=0}^{\infty} \frac{(2x-1)^n}{5^n}$.

**Step 1. Set up the ratio.** The ratio of consecutive terms is $\frac{|2x - 1|}{5}$, with no $n$ left over, so $L = \frac{|2x-1|}{5}$.

**Step 2. Solve $L < 1$.** $|2x - 1| < 5$. Factor out the 2 to put it in the form $|x - a| < R$:

$$2\left|x - \tfrac{1}{2}\right| < 5 \quad\Longrightarrow\quad \left|x - \tfrac{1}{2}\right| < \tfrac{5}{2}$$

So the center is $a = \frac{1}{2}$ and the radius is $R = \frac{5}{2}$, not 5. The open interval is $-2 < x < 3$.

**Step 3. Check the endpoints.** At $x = 3$ the series is $\sum 1^n$, and at $x = -2$ it is $\sum (-1)^n$. Neither has terms going to 0, so both diverge. The interval is $(-2, 3)$.

## Common mistakes with power series

1. **Forgetting the endpoints.** This is the most common lost point. The ratio test never decides them.
2. **Reading off the wrong center.** $(x + 2)^n$ is centered at $-2$, and $(3x - 6)^n$ is centered at $2$.
3. **Not factoring out the coefficient of $x$.** From $|2x - 1| < 5$ the radius is $\frac{5}{2}$. Always rewrite as $|x - a| < R$ first.
4. **Dropping the absolute value.** The ratio test uses $|x - a|$. Without it you lose the left half of the interval.
5. **Letting $x$ "go to infinity" with $n$.** In the limit, $x$ is a fixed number. Only $n$ changes.
6. **Using the ratio test result at an endpoint.** At an endpoint you need a different test, applied to the series with the number substituted in.

## Quick reference: radius and interval of convergence

| Series | Ratio test limit $L$ | Radius $R$ | Interval |
| --- | --- | --- | --- |
| $\sum x^n$ | $\lvert x \rvert$ | $1$ | $(-1, 1)$ |
| $\sum \frac{x^n}{n}$ | $\lvert x \rvert$ | $1$ | $[-1, 1)$ |
| $\sum \frac{x^n}{n^2}$ | $\lvert x \rvert$ | $1$ | $[-1, 1]$ |
| $\sum \frac{(x-3)^n}{n^2\, 2^n}$ | $\frac{\lvert x-3 \rvert}{2}$ | $2$ | $[1, 5]$ |
| $\sum \frac{x^n}{n!}$ | $0$ | $\infty$ | $(-\infty, \infty)$ |
| $\sum n!\, x^n$ | $\infty$ for $x \ne 0$ | $0$ | $\{0\}$ |

A useful pattern from the first three rows: polynomial factors like $n$ or $n^2$ do not change the radius, but they can change what happens at the endpoints.

## Practice until the steps feel automatic

Finding an interval of convergence is a fixed routine: ratio, limit, solve, check two endpoints. The routine only becomes fast after you have run it on a dozen different series. The interactive lesson below gives you fresh series to work through, checks each step instantly, and shows you exactly where an endpoint test went wrong.

## Frequently asked questions

### What is the radius of convergence of a power series?

It is the number $R$ such that the series $\sum c_n (x-a)^n$ converges when $|x-a| < R$ and diverges when $|x-a| > R$. It measures how far from the center $a$ you can go before the series stops working. $R$ can be $0$, a positive number, or infinite.

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

The radius is a single number, the distance from the center to the edge. The interval is the actual set of $x$ values where the series converges, including any endpoints that work. Two series can share the radius $R = 1$ but have intervals $(-1, 1)$, $[-1, 1)$ or $[-1, 1]$.

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

At the endpoints the ratio test limit equals exactly 1, and a ratio test result of 1 is inconclusive. So you substitute each endpoint into the series and use another test, such as the p-series test, the alternating series test or the nth-term test.

### Can a power series converge for every x?

Yes. If the ratio test limit is 0 for every $x$, the radius is infinite and the interval is $(-\infty, \infty)$. The series for $e^x$, $\sin x$ and $\cos x$ all behave this way, usually because of a factorial in the denominator.

### Does a power series always converge at its center?

Yes. At $x = a$ every term after the first is zero, so the series equals $c_0$. That is why the smallest possible interval of convergence is the single point $\{a\}$, which happens when $R = 0$.