Interval of Convergence

From a Radius to an Interval

The Radius of Convergence tutorial showed that a power series converges absolutely at every point strictly inside its radius and diverges at every point strictly outside it. The remaining question is what happens at the boundary. When the radius is finite and positive, there are two boundary points, one on each side of the center. The series must be tested at each of them; neither the radius nor the behavior at one endpoint always decides the other.

The interval of convergence is the set of real inputs where the power series converges. The radius determines the open interval that is guaranteed to be inside this set. Endpoint tests determine whether either end of that interval is included. Together, these facts describe the entire convergence set on the real line.

Definition and Classification

Definition: The interval of convergence of a real power series \(\sum_{n=0}^{\infty}c_n(x-a)^n\) is the set of all real numbers \(x\) for which the series converges. Convergence here means ordinary convergence; it need not be absolute.

Suppose first that the radius \(R\) is finite and positive. The two boundary points are \(a-R\) and \(a+R\). At every point strictly between them, the series converges absolutely by the Theorem (Convergence Inside and Outside the Radius). At every point outside the closed interval between them, it diverges by the same theorem. Only the two boundary points require additional tests.

Theorem (Endpoint Classification of the Interval of Convergence): Let \(R\) be the radius of convergence of \(\sum_{n=0}^{\infty}c_n(x-a)^n\).

If \(0<R<\infty\), the series converges at every \(x\) with \(a-R<x<a+R\), diverges at every \(x\) with \(|x-a|>R\), and its convergence set is obtained by adding to \((a-R,a+R)\) whichever of \(a-R\) and \(a+R\) are points of convergence.
If \(R=0\), its convergence set is \(\{a\}\).
If \(R=\infty\), its convergence set is all of \(\mathbb{R}\).

Proof. First suppose \(0<R<\infty\). If \(a-R<x<a+R\), then \(|x-a|<R\), so the series converges absolutely by the Theorem (Convergence Inside and Outside the Radius). If \(|x-a|>R\), that theorem says the series diverges. Every real \(x\) not covered by these two cases must satisfy \(|x-a|=R\), which means \(x=a-R\) or \(x=a+R\). The convergence set therefore consists of the open interval together with exactly those endpoints where the series converges.

If \(R=0\), the series converges at \(a\), because every term of positive degree vanishes there and only the constant term may remain. For every \(x\neq a\), \(|x-a|>0=R\), so the series diverges by the Theorem (Convergence Inside and Outside the Radius). Its convergence set is therefore \(\{a\}\).

If \(R=\infty\), every real \(x\) satisfies \(|x-a|<R\), so the series converges absolutely at every real input. This proves all three cases. \(\square\)

For a finite positive radius, the theorem gives four possible intervals: neither endpoint included, just the left included, just the right included, or both included. Endpoint convergence is ordinary convergence, so an endpoint can belong to the interval even when the series is not absolutely convergent there.

Testing the Boundary Points

At the right endpoint, substitute \(x=a+R\) into the original series. At the left endpoint, substitute \(x=a-R\). These substitutions produce two numerical series, which may have different signs and different convergence behavior. Apply an appropriate series test to each one, using the results established earlier in this course.

Find the radius and center.
Use the coefficient formula or another established method to determine \(R\) and \(a\).

Write the candidate interval.
When \(0<R<\infty\), begin with \((a-R,a+R)\). The interior is absolutely convergent.

Substitute the left endpoint.
Set \(x=a-R\) in the series and determine whether the resulting numerical series converges.

Substitute the right endpoint.
Set \(x=a+R\) and test the resulting series separately. Include each endpoint exactly when its series converges.

The two endpoint series have an important connection when absolute convergence is considered. Their term magnitudes agree, even though their signs may differ.

Theorem (Absolute Endpoint Convergence Is Symmetric): Suppose \(0<R<\infty\) is the radius of \(\sum_{n=0}^{\infty}c_n(x-a)^n\). The series converges absolutely at \(a-R\) if and only if it converges absolutely at \(a+R\).

Proof. At the right endpoint, the \(n\)th term is \(c_nR^n\), so the series of absolute values is \(\sum_{n=0}^{\infty}|c_n|R^n\). At the left endpoint, the \(n\)th term is \(c_n(-R)^n\), and its absolute value is

|c_n(-R)^n|=|c_n|\,|-R|^n=|c_n|R^n.

Thus the two series of absolute values have identical terms for every \(n\). One converges exactly when the other does, proving the claim. \(\square\)

This symmetry concerns absolute convergence, not ordinary convergence. The signs at the endpoints can differ, allowing one endpoint series to converge conditionally while the other diverges. Endpoint tests must therefore remain separate unless absolute convergence has been established.

Worked Examples: Including the Endpoints

Worked Example: One Endpoint Included

Consider the power series centered at \(2\):

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

Its coefficients are \(c_n=1/n\) for \(n\geq1\), with constant coefficient \(c_0=0\). Since \((1/n)^{1/n}\to1\), the Cauchy–Hadamard Formula gives \(R=1\). The interior interval is therefore \((1,3)\).

At the left endpoint \(x=1\), we have \(x-2=-1\), giving

\sum_{n=1}^{\infty}\frac{(-1)^n}{n}.

This is an alternating series with terms of magnitude \(1/n\). These magnitudes decrease to zero, so the Alternating Series Test proves convergence. It is not absolutely convergent, because the absolute-value series is \(\sum_{n=1}^{\infty}1/n\), which diverges by the \(p\)-Series Convergence Criterion with \(p=1\). Thus \(x=1\) is included.

At the right endpoint \(x=3\), we have \(x-2=1\), giving \(\sum_{n=1}^{\infty}1/n\), which diverges. The interval of convergence is \([1,3)\).

Worked Example: Both Endpoints Included Absolutely

Consider the series centered at \(-1\):

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

Here \(c_n=1/n^2\) for \(n\geq1\), and \((1/n^2)^{1/n}\to1\), so the radius is \(R=1\). The candidate interior is \((-2,0)\). At either endpoint, the absolute values of the terms are \(1/n^2\), and

\sum_{n=1}^{\infty}\left|\frac{(\pm1)^n}{n^2}\right| = \sum_{n=1}^{\infty}\frac{1}{n^2}.

The \(p\)-Series Convergence Criterion, with \(p=2\), shows that this series converges. Thus both endpoints are included, and the interval of convergence is \([-2,0]\). In this example the endpoint convergence is absolute on both sides, as the Absolute Endpoint Convergence Is Symmetric theorem predicts.

Worked Example: Neither Endpoint Included

For the geometric power series centered at \(4\),

\sum_{n=0}^{\infty}(x-4)^n,

the coefficients are all \(1\), so the Cauchy–Hadamard Formula gives \(R=1\). The interior interval is \((3,5)\). At \(x=3\), the terms are \((-1)^n\); at \(x=5\), they are \(1\). In both cases the terms fail to tend to zero: their absolute values equal \(1\) for every \(n\). By the Necessary Condition for Series Convergence, the series diverges at both endpoints. Its interval of convergence is therefore \((3,5)\).

Worked Example: The Other Single-Endpoint Pattern

Consider

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

The coefficient magnitudes are \(1/n\), so the radius is \(1\), and the candidate interior is \((1,3)\). At \(x=3\), the terms are \((-1)^n/n\), which form a convergent alternating series by the Alternating Series Test. At \(x=1\), the terms are \(1/n\), so the series diverges by the \(p\)-Series Convergence Criterion. Hence the interval is \((1,3]\). Compared with the first example, the endpoint behavior has switched sides.

Why Endpoint Tests Matter

The examples show why it is not enough to report only the radius. Each series has radius \(1\), but their intervals of convergence differ: \([1,3)\), \([-2,0]\), \((3,5)\), and \((1,3]\). The radius gives the distance to the boundary; it does not say which boundary points converge. The endpoint substitutions reveal information that the coefficient growth rate alone may leave undecided.

A common pitfall is to write a closed interval automatically because the radius is a finite number. The interior is guaranteed, but the endpoints are not. Another is to assume the two endpoints behave alike because they are the same distance from the center. Their absolute-value series are identical, but their ordinary series may differ because the signs change. Test both endpoints and distinguish absolute convergence from conditional convergence.

Takeaway: For a finite positive radius \(R\), start with \((a-R,a+R)\), then test \(a-R\) and \(a+R\) separately and include each convergent endpoint. Absolute convergence at either endpoint occurs at both; ordinary convergence need not.

Check Your Understanding

Use the radius theorem and the endpoint tests to answer the following questions.

For a power series of finite positive radius \(R\) centered at \(a\), which points are guaranteed to be in the interval of convergence?
Why must the two boundary points be tested separately for ordinary convergence?
If the series converges absolutely at \(a-R\), what can you conclude about absolute convergence at \(a+R\)?
A power series has radius \(2\) and center \(5\). What is its guaranteed open interval of convergence, and which two inputs need endpoint tests?
What is the convergence set when the radius is zero? What is it when the radius is infinite?

What You'll Learn