A Complete Endpoint-Analysis Workshop

A Reliable Endpoint-Analysis Workflow

A power series can converge throughout the interior of a certain interval and behave differently at its boundary points. Finding the radius identifies where to investigate; it does not, by itself, decide whether either endpoint belongs in the interval of convergence.

This workshop puts the full process together: find the radius, substitute each endpoint into the original series, apply a suitable convergence test, and report the exact interval. The endpoint tests may involve familiar series such as the harmonic series, an alternating series, or a telescoping series.

Definition: A power series centered at \(a\) has the form \(\sum_{n=0}^{\infty}c_n(x-a)^n\), where the coefficients \(c_n\) are fixed. Its interval of convergence is the set of real numbers \(x\) for which the series converges.

First Find the Radius

For the positive-degree terms, the root test is governed by the limsup of the coefficient roots. Define \(L=\limsup_{n\to\infty}|c_n|^{1/n}\). The radius \(R\) is given by the following conventions:

R=\frac{1}{L},\qquad \frac{1}{0}=\infty,\qquad \frac{1}{\infty}=0.

When \(0<L<\infty\), the root test gives absolute convergence if \(|x-a|<R\) and divergence if \(|x-a|>R\). The same interior and exterior conclusions hold in the limiting cases \(L=0\) and \(L=\infty\). When \(0<R<\infty\), the root test does not settle convergence at \(|x-a|=R\): each endpoint must be checked using its own series. If \(R=0\), the series converges only at the center; if \(R=\infty\), there are no finite boundary points.

Why the radius formula works: For \(x\ne a\), the nth root of the absolute value of the \(n\)th positive-degree term is \(|c_n|^{1/n}|x-a|\). Because \(|x-a|\) is a fixed positive number, the limsup of this expression is \(L|x-a|\). The root test gives convergence when this is less than \(1\), and divergence when it is greater than \(1\), which yields the radius \(R=1/L\). At \(x=a\), this calculation is not used: for every \(n\ge1\), \(c_n(x-a)^n=0\), so the series converges there. This separate treatment also applies if \(L=\infty\), for which the expression \(L|x-a|\) at the center would be undefined.

The constant term \(c_0\) has no effect on convergence. When the radius is finite and positive, the only undecided points are \(a-R\) and \(a+R\). Substitution at each point produces a numerical series, which is then tested on its own.

Four Complete Examples

Worked Example: Polynomial Coefficients

Find the exact interval of convergence of \(\sum_{n=1}^{\infty}n^2\left(\frac{x-1}{3}\right)^n\).

Here the center is \(a=1\), and the coefficient of \((x-1)^n\) is \(c_n=n^2/3^n\). Therefore

|c_n|^{1/n}=\frac{(n^2)^{1/n}}{3}\longrightarrow\frac13, \qquad R=3.

The interior \(|x-1|<3\) converges absolutely. At \(x=4\), the series becomes \(\sum_{n=1}^{\infty}n^2\); its terms \(n^2\) do not tend to zero, so it diverges. At \(x=-2\), it becomes \(\sum_{n=1}^{\infty}n^2(-1)^n\); the absolute values of its terms are \(n^2\), which also do not tend to zero, so it diverges.

Both endpoints are excluded, giving the interval \((-2,4)\). As a check, the endpoints are \(1-3=-2\) and \(1+3=4\), as required by the center and radius.

Worked Example: A Logarithmic Boundary Series

Find the exact interval of convergence of \(\sum_{n=2}^{\infty}\frac{x^n}{n\ln n}\).

The coefficient roots satisfy \(\left(\frac1{n\ln n}\right)^{1/n}\to1\), so \(R=1\). The series converges absolutely for \(|x|<1\) and diverges for \(|x|>1\).

At \(x=1\), it becomes \(\sum_{n=2}^{\infty}\frac1{n\ln n}\). The function \(1/(t\ln t)\) is positive and decreasing for \(t\ge2\), and

\int_2^B\frac{dt}{t\ln t} =\ln(\ln B)-\ln(\ln 2)\longrightarrow\infty \quad\text{as }B\to\infty.

The integral test therefore gives divergence at \(x=1\). At \(x=-1\), the series is \(\sum_{n=2}^{\infty}\frac{(-1)^n}{n\ln n}\). Its positive term magnitudes decrease to zero, so the alternating-series test gives convergence. It is not absolute convergence, because the absolute-value series is the divergent series just tested at \(x=1\). Thus the convergence at \(-1\) is conditional. The exact interval is \([-1,1)\).

As a check, \(-1\) is included and \(1\) excluded; the series is absolutely convergent between them and outside neither boundary test was assumed.

Worked Example: Alternating Coefficients

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

The coefficient roots are \(\left|(-1)^{n-1}/n\right|^{1/n}=n^{-1/n}\to1\). Hence the center is \(2\), the radius is \(1\), and the series converges absolutely for \(1<x<3\).

At the right endpoint \(x=3\), substitution gives \(\sum_{n=1}^{\infty}(-1)^{n-1}/n\), the alternating harmonic series. The terms decrease in magnitude to zero, so it converges; its absolute-value series is harmonic and diverges. Thus convergence at \(3\) is conditional.

At the left endpoint \(x=1\), each term is

(-1)^{n-1}\frac{(1-2)^n}{n} =(-1)^{n-1}\frac{(-1)^n}{n} =-\frac1n.

The endpoint series is \(-\sum_{n=1}^{\infty}1/n\), which diverges. Therefore the exact interval is \((1,3]\). The endpoints are one unit from the center \(2\), confirming the radius-based boundary values.

Worked Example: Telescoping at One Endpoint

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

Since \(\left(\frac1{n(n+1)}\right)^{1/n}\to1\), the radius is \(1\). The series converges absolutely for \(|x|<1\).

At \(x=1\), use \(\frac1{n(n+1)}=\frac1n-\frac1{n+1}\). The \(N\)th partial sum is

\sum_{n=1}^{N}\frac1{n(n+1)} =\sum_{n=1}^{N}\left(\frac1n-\frac1{n+1}\right) =1-\frac1{N+1}\longrightarrow1.

Thus the series converges at \(1\), with sum \(1\). At \(x=-1\), the series is \(\sum_{n=1}^{\infty}\frac{(-1)^n}{n(n+1)}\). Its absolute-value series is the telescoping series just evaluated, so it converges absolutely. Both endpoints are included, giving \([-1,1]\).

As a check, the absolute values at \(-1\) have partial sums \(1-\frac1{N+1}\), bounded above by \(1\); this confirms the claimed absolute convergence there.

What Endpoint Results Do—and Do Not—Tell You

At a boundary point, the series must be tested after substituting the endpoint. A radius calculation tells you where to perform this test, not its result. The same radius can accompany different endpoint behavior: one endpoint may diverge, the other may converge conditionally, or both may converge absolutely.

At minimum, verify that the terms of each endpoint series tend to zero. If they do not, the series diverges. If they do, a stronger test may be needed: comparison or limit comparison for positive terms, the alternating-series test for suitable alternating terms, or telescoping when partial cancellation occurs. In particular, alternating signs alone do not ensure convergence; the term magnitudes must also satisfy the conditions of the test being used.

Takeaway: A complete interval report includes the radius and a separate convergence verdict at every boundary point. State whether an included endpoint is absolute or conditional when that distinction applies.

A Common Boundary Mistake

A common error is to use the root test’s radius conclusion as if it decided convergence at the endpoints. For example, finding \(R=1\) for \(\sum_{n=1}^{\infty}x^n/n\) does not justify writing \([-1,1]\). That conclusion includes \(x=1\), where substitution gives the divergent harmonic series \(\sum 1/n\).

The correction is to test both endpoints. At \(x=-1\), the series is \(\sum(-1)^n/n\), which converges conditionally by the alternating-series test. At \(x=1\), it diverges. The exact interval is therefore \([-1,1)\). The root test establishes convergence inside the radius and divergence outside it; the endpoint series supply the remaining information.

Summary of the Procedure

Identify the center and coefficients.
Write the series as \(\sum c_n(x-a)^n\), and note which powers are present.

Calculate the radius.
Find \(L=\limsup |c_n|^{1/n}\) and use \(R=1/L\), including the conventions for \(L=0\) or \(L=\infty\).

Establish the interior and exterior behavior.
The series converges absolutely for \(|x-a|<R\) and diverges for \(|x-a|>R\).

Test the boundary points separately.
When \(0<R<\infty\), substitute \(x=a-R\) and \(x=a+R\) into the original series, then use an appropriate test on each resulting series. If \(R=0\) or \(R=\infty\), there are no boundary points to test.

Write the exact interval.
Include precisely the endpoints where the series converges, and identify conditional convergence when relevant.

Check Your Understanding

For each question, explain the endpoint reasoning as well as the radius calculation where one is requested.

What does the root test establish when \(|x-a|=R\)?
Why must \(x=a\) be handled separately when using the limsup radius formula?
For \(\sum_{n=1}^{\infty}x^n/n^2\), find the radius and test both endpoints. What is the exact interval?
For \(\sum_{n=1}^{\infty}(-1)^{n-1}(x-1)^n/\sqrt n\), find the radius and determine what happens at each endpoint.
What is the difference between conditional and absolute convergence at an endpoint?