Interval of Convergence: Radius & Examples

The radius of convergence R of a power series $\sum c_n(x - a)^n$ is how far from the center a the series converges. The interval of convergence runs from a − R to a + R, plus any endpoint where the series also converges. Find R with the ratio test, then test each endpoint separately.

Before this lesson: Power Series, Ratio Test, Alternating Series Test and p-Series

Why a power series converges on an interval

A power series is like a geometric series whose ratio depends on x. Close to the center, the terms shrink fast and the series converges. Far from the center, the terms grow and it diverges. The switch happens at the same distance R on both sides of the center.

Inside the interval, the series converges absolutely. Outside, it diverges. At the two endpoints, anything can happen, so you must test them one at a time.

Number line from −3 to 7 for the series of (x − 2)ⁿ/(n · 3ⁿ). The center is 2 and the radius is 3. A thick blue segment runs from a closed dot at −1 (included) to an open circle at 5 (not included), labeled converges for −1 ≤ x < 5. The parts of the line left of −1 and right of 5 are labeled diverges. — The series in Example 3 converges within 3 units of its center, 2, including the endpoint −1 but not the endpoint 5.

How to find the radius and interval of convergence

Apply the ratio test to the absolute values, keeping x as a variable: $L = \lim_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n}\right|$ .
Set L < 1 and solve for |x − a| < R. That number R is the radius of convergence.
Write the open interval a − R < x < a + R.
Put each endpoint into the series. Test each resulting number series with another test, such as a p-series or the alternating series test.
Include each endpoint where the series converges.

Three possible cases

Case	What the ratio test gives	Interval of convergence
R = 0	L = ∞ for every x ≠ a	Only the center, x = a
R is a positive number	L < 1 exactly when \|x − a\| < R	From a − R to a + R, with endpoints checked
R = ∞	L = 0 for every x	All real numbers

Interval of convergence examples

Example 1: Converges for every x

Find the radius and interval of convergence of $\sum_{n=0}^{\infty} \dfrac{x^n}{n!}$ .

\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{|x|^{n+1}}{(n + 1)!} \cdot \dfrac{n!}{|x|^n}

Use the ratio test with x as a variable. Multiply by the reciprocal of |aₙ|.

\dfrac{|x|^{n+1}}{(n + 1)!} \cdot \dfrac{n!}{|x|^n} = \dfrac{|x|}{n + 1}

Cancel: (n + 1)! = (n + 1) · n!.

L = \lim_{n \to \infty} \dfrac{|x|}{n + 1} = 0

For any fixed x, the bottom grows without bound.

0 \lt 1 \text{ for every } x

The condition L < 1 always holds.

Answer: R = ∞, and the interval of convergence is all real numbers. (This series equals $e^x$ .)

Example 2: Converges only at the center

Find the radius of convergence of $\sum_{n=0}^{\infty} n!\,x^n$ .

\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{(n + 1)!\,|x|^{n+1}}{n!\,|x|^n}

Use the ratio test, for x ≠ 0.

\dfrac{(n + 1)!\,|x|^{n+1}}{n!\,|x|^n} = (n + 1)|x|

Cancel n! and |x|ⁿ.

L = \lim_{n \to \infty} (n + 1)|x| = \infty

Any nonzero |x| times a growing n + 1 grows without bound.

Answer: R = 0. The series converges only at its center, x = 0.

Example 3: A finite radius with endpoints to check

Find the radius and interval of convergence of $\sum_{n=1}^{\infty} \dfrac{(x - 2)^n}{n \cdot 3^n}$ .

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

Use the ratio test with x as a variable.

\dfrac{|x - 2|^{n+1}}{(n + 1)3^{n+1}} \cdot \dfrac{n \cdot 3^n}{|x - 2|^n} = \dfrac{|x - 2|}{3} \cdot \dfrac{n}{n + 1}

Cancel the powers (see exponent rules).

L = \dfrac{|x - 2|}{3}

n/(n + 1) goes to 1.

\dfrac{|x - 2|}{3} \lt 1 \quad \Leftrightarrow \quad |x - 2| \lt 3

Set L < 1 and multiply by 3. So R = 3.

-1 \lt x \lt 5

Solve the absolute value inequality: 2 − 3 < x < 2 + 3.

\text{At } x = 5: \quad \sum_{n=1}^{\infty} \dfrac{3^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \dfrac{1}{n}

This is the harmonic series, which diverges. Leave out x = 5.

\text{At } x = -1: \quad \sum_{n=1}^{\infty} \dfrac{(-3)^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n}

This converges by the alternating series test. Include x = −1.

Answer: R = 3, and the interval of convergence is −1 ≤ x < 5, written [−1, 5).

Example 4: Both endpoints included

Find the interval of convergence of $\sum_{n=1}^{\infty} \dfrac{x^n}{n^2}$ .

\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{|x|^{n+1}}{(n + 1)^2} \cdot \dfrac{n^2}{|x|^n} = |x|\left(\dfrac{n}{n + 1}\right)^2

Use the ratio test and cancel.

L = |x| \lt 1

(n/(n + 1))² goes to 1. So R = 1.

\text{At } x = 1: \quad \sum_{n=1}^{\infty} \dfrac{1}{n^2}

A p-series with p = 2 > 1, so it converges.

\text{At } x = -1: \quad \sum_{n=1}^{\infty} \dfrac{(-1)^n}{n^2}

Its absolute values form the same p-series, so it converges absolutely.

Answer: R = 1, and the interval of convergence is −1 ≤ x ≤ 1, written [−1, 1].

Example 5: Where is a model valid? (word problem)

A signal model uses $f(t) = \sum_{n=0}^{\infty} \dfrac{(n + 1)t^n}{4^n}$ . For which values of t does the model give a finite value?

\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{(n + 2)|t|^{n+1}}{4^{n+1}} \cdot \dfrac{4^n}{(n + 1)|t|^n}

Use the ratio test with t as the variable.

\dfrac{(n + 2)|t|^{n+1}}{4^{n+1}} \cdot \dfrac{4^n}{(n + 1)|t|^n} = \dfrac{|t|}{4} \cdot \dfrac{n + 2}{n + 1}

Cancel the powers.

L = \dfrac{|t|}{4} \lt 1 \quad \Leftrightarrow \quad |t| \lt 4

(n + 2)/(n + 1) goes to 1. So R = 4.

\text{At } t = 4: \quad \sum (n + 1) = 1 + 2 + 3 + \cdots

The terms grow, so it diverges by the divergence test.

\text{At } t = -4: \quad \sum (-1)^n (n + 1) = 1 - 2 + 3 - \cdots

The terms do not go to 0, so it diverges too.

Answer: the model is valid only for −4 < t < 4.

Common mistakes

Mistake	Example of the error	Fix
Skipping the endpoints	Giving (−1, 5) for Example 3	The ratio test says nothing at L = 1. Test each endpoint in the original series.
Using the ratio test at an endpoint	Getting L = 1 at x = 5 and calling it convergent	At an endpoint L is always 1. Use a p-series, alternating series or comparison test.
Forgetting the center	Writing \|x\| < 3 for a series in powers of (x − 2)	Keep x − a: \|x − 2\| < 3 gives −1 < x < 5.
Mixing up R and the interval	Saying “R = [−1, 5)”	R is a distance (3). The interval is a set of x values.
Missing a coefficient on x	For $\sum (2x - 1)^n$ , saying R = 1	\|2x − 1\| < 1 means \|x − 1/2\| < 1/2, so R = 1/2.
Dropping the absolute value	Writing (x − 2)/3 < 1, so x < 5	Use \|x − 2\|/3 < 1, which bounds x on both sides.

Related methods

Using the root test

If the terms are nth powers, such as $\left(\frac{x}{n}\right)^n$ , the root test finds L more easily. Set the limit of $\sqrt[n]{|a_n|}$ less than 1, exactly as with the ratio test.

Geometric series shortcut

For a geometric power series like $\sum \left(\frac{x}{5}\right)^n$ , you can skip the ratio test: it converges exactly when |ratio| < 1, and both endpoints always diverge.

Practice problems

Find the interval of convergence of $\sum_{n=0}^{\infty} \dfrac{x^n}{5^n}$ .
Find the interval of convergence of $\sum_{n=1}^{\infty} \dfrac{x^n}{n}$ .
Find the radius and interval of convergence of $\sum_{n=1}^{\infty} \dfrac{(x + 1)^n}{n^2 \cdot 2^n}$ .
Find the radius of convergence of $\sum_{n=0}^{\infty} \dfrac{3^n x^n}{n!}$ .
Find the radius of convergence of $\sum_{n=1}^{\infty} n^n x^n$ .
Find the radius and interval of convergence of $\sum_{n=1}^{\infty} \dfrac{(2x - 1)^n}{\sqrt{n}}$ .
Find the interval of convergence of $\sum_{n=0}^{\infty} \dfrac{x^{2n}}{4^n}$ .
Find the interval of convergence of $\sum_{n=1}^{\infty} n(x - 3)^n$ .
A power series centered at 2 converges at x = 6 and diverges at x = −4. What can you say about R?
A model uses $\sum_{n=1}^{\infty} \dfrac{v^n}{n \cdot 10^n}$ . For which values of v does it converge?

Show answers

\sum_{n=0}^{\infty} \left(\dfrac{x}{5}\right)^n \quad \text{converges when } \left|\dfrac{x}{5}\right| \lt 1

It is geometric with ratio x/5.

|x| \lt 5

Multiply by 5. So R = 5.

\text{At } x = \pm 5: \quad \sum 1^n \text{ or } \sum (-1)^n

The terms do not go to 0, so both endpoints diverge.

Answer: −5 < x < 5

L = \lim_{n \to \infty} |x| \cdot \dfrac{n}{n + 1} = |x|

Ratio test; R = 1.

\text{At } x = 1: \quad \sum \dfrac{1}{n}

Harmonic series: diverges.

\text{At } x = -1: \quad \sum \dfrac{(-1)^n}{n}

Converges by the alternating series test.

Answer: −1 ≤ x < 1

L = \lim_{n \to \infty} \dfrac{|x + 1|}{2}\left(\dfrac{n}{n + 1}\right)^2 = \dfrac{|x + 1|}{2}

Ratio test.

|x + 1| \lt 2 \quad \Leftrightarrow \quad -3 \lt x \lt 1

So R = 2 and the center is −1.

\text{At } x = 1: \; \sum \dfrac{1}{n^2}; \quad \text{at } x = -3: \; \sum \dfrac{(-1)^n}{n^2}

Both converge (p-series with p = 2, and its alternating version).

Answer: R = 2; −3 ≤ x ≤ 1

\left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{3^{n+1}|x|^{n+1}}{(n + 1)!} \cdot \dfrac{n!}{3^n |x|^n} = \dfrac{3|x|}{n + 1}

Ratio test and cancel.

L = 0 \lt 1 \text{ for every } x

The bottom grows without bound.

Answer: R = ∞ (all real numbers)

\sqrt[n]{|n^n x^n|} = n|x|

The terms are nth powers, so use the root test.

\lim_{n \to \infty} n|x| = \infty \text{ for } x \ne 0

It is less than 1 only at x = 0.

Answer: R = 0 (converges only at x = 0)

L = \lim_{n \to \infty} |2x - 1|\sqrt{\dfrac{n}{n + 1}} = |2x - 1|

Ratio test.

|2x - 1| \lt 1 \quad \Leftrightarrow \quad 0 \lt x \lt 1

|x − 1/2| < 1/2, so R = 1/2.

\text{At } x = 1: \; \sum \dfrac{1}{\sqrt{n}}; \quad \text{at } x = 0: \; \sum \dfrac{(-1)^n}{\sqrt{n}}

The first is a p-series with p = 1/2 (diverges); the second converges by the alternating series test.

Answer: R = 1/2; 0 ≤ x < 1

\sum_{n=0}^{\infty} \left(\dfrac{x^2}{4}\right)^n \quad \text{converges when } \dfrac{x^2}{4} \lt 1

Geometric with ratio x²/4.

x^2 \lt 4 \quad \Leftrightarrow \quad |x| \lt 2

So R = 2.

\text{At } x = \pm 2: \quad \sum 1

The terms are all 1, so it diverges.

Answer: −2 < x < 2

L = \lim_{n \to \infty} |x - 3| \cdot \dfrac{n + 1}{n} = |x - 3|

Ratio test; R = 1, center 3.

\text{At } x = 4 \text{ and } x = 2: \quad \sum n, \; \sum (-1)^n n

The terms do not go to 0, so both diverge.

Answer: 2 < x < 4

|6 - 2| = 4 \le R

It converges at a distance of 4 from the center, so R is at least 4.

|-4 - 2| = 6 \ge R

It diverges at a distance of 6, so R is at most 6.

Answer: 4 ≤ R ≤ 6

10.

L = \lim_{n \to \infty} \dfrac{|v|}{10} \cdot \dfrac{n}{n + 1} = \dfrac{|v|}{10}

Ratio test; R = 10.

\text{At } v = 10: \; \sum \dfrac{1}{n}; \quad \text{at } v = -10: \; \sum \dfrac{(-1)^n}{n}

The first diverges; the second converges.

Answer: −10 ≤ v < 10

Interval of convergence FAQ

What is the radius of convergence?

It is the distance R from the center within which a power series converges. The series converges absolutely when |x − a| < R and diverges when |x − a| > R.

How do you find the interval of convergence?

Use the ratio test with x as a variable, set the limit less than 1 to find R, then test each endpoint separately in the original series.

Why do you have to check the endpoints?

At an endpoint the ratio test gives L = 1, which tells you nothing. The series might converge at both endpoints, one, or neither.

Can the radius of convergence be 0 or infinity?

Yes. If the series converges only at its center, R = 0. If it converges for every x, like the series for $e^x$ , R = ∞.

Summary

Use the ratio test with x as a variable, and set L < 1.
Solve for |x − a| < R to get the radius of convergence R.
Test both endpoints with other tests; include those that converge.
R can be 0 (only the center), a positive number, or ∞ (every x).
Inside the interval the series converges absolutely; outside it diverges.

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