Testing the Two Endpoints Independently

Why the Endpoints Need Separate Tests

A power series with a finite radius of convergence has two boundary points, one on each side of its center. The radius describes where the series converges in the interior and diverges outside, but it does not determine what happens at either boundary point.

At the boundaries, substituting the values of \(x\) produces two ordinary numerical series. Their terms can have different signs, even though the boundary points are equally far from the center. Each resulting series must therefore be tested on its own.

Key principle: A power series may converge at one endpoint and diverge at the other. The radius of convergence alone cannot decide either endpoint.

Substitute Each Endpoint Carefully

Consider a power series centered at \(c\), with finite positive radius of convergence \(R\):

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

The boundary points are \(c-R\) and \(c+R\). At the left endpoint, \(x-c=-R\); at the right endpoint, \(x-c=R\). Direct substitution gives:

\begin{aligned} x=c-R &: \quad \sum_{n=0}^{\infty}a_n(-R)^n,\\ x=c+R &: \quad \sum_{n=0}^{\infty}a_nR^n. \end{aligned}

The factor \((-R)^n=(-1)^nR^n\) can change the signs of the terms at the left endpoint. At the right endpoint, the factor is \(R^n\), so the coefficients keep their original signs. This difference is often decisive.

Endpoint procedure: Substitute the left endpoint and simplify its series; test it. Then substitute the right endpoint and simplify its series; test it separately. Do not infer one result from the other.

A Useful Pair of Endpoint Series

Several examples below reduce to one of two series: the alternating series \(\sum_{n=0}^{\infty}(-1)^n/(n+1)\), or the positive series \(\sum_{n=0}^{\infty}1/(n+1)\). Here is a direct justification of their different behavior.

For the alternating series, let \(S_N=\sum_{n=0}^{N}(-1)^n/(n+1)\). Grouping consecutive terms gives the partial sums with even and odd final indices. For \(m\geq 0\),

\begin{aligned} S_{2m+2}-S_{2m} &=-\frac{1}{2m+2}+\frac{1}{2m+3}<0,\\ S_{2m+3}-S_{2m+1} &=\frac{1}{2m+3}-\frac{1}{2m+4}>0. \end{aligned}

Thus the even-indexed partial sums decrease, while the odd-indexed partial sums increase. For every \(m\geq 0\), the two neighboring partial sums satisfy

S_{2m+1}\leq S_{2m+2} \qquad\text{and}\qquad S_{2m+2}-S_{2m+1}=\frac{1}{2m+3}.

The increasing sequence \(S_{2m+1}\) is bounded above by \(S_0=1\), and the decreasing sequence \(S_{2m+2}\) is bounded below by \(S_1=1/2\). Both therefore have limits. Their difference tends to zero, so those limits are equal. The full sequence of partial sums converges to that common limit.

In contrast, the positive series diverges. To see this, group its terms into blocks. For each \(k\geq 0\), the block with denominators \(2^k+1\) through \(2^{k+1}\) contains \(2^k\) terms, each at least \(1/2^{k+1}\). Its sum is therefore at least \(1/2\). As the number of such blocks increases, the partial sums grow without bound.

What the comparison shows: These two endpoint series differ only by alternating signs, yet one converges and the other diverges. The endpoint signs cannot be ignored.

Worked Examples

Worked Example: Convergence at the Left Endpoint

Find the interval of convergence of

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

First, the ratio of successive term magnitudes, when \(x\neq 1\), is

\frac{|x-1|^{n+1}/(n+2)}{|x-1|^n/(n+1)} = |x-1|\frac{n+1}{n+2} \longrightarrow |x-1|.

The ratio test gives convergence when \(|x-1|<1\), so the center is \(1\), the radius is \(1\), and the boundary points are \(0\) and \(2\). Now test them separately.

At \(x=0\), \(x-1=-1\), so substitution gives

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

which converges by the partial-sum argument above. Substituting \(x=0\) into the original power gives exactly the factor \((-1)^n\), confirming the endpoint series.

At \(x=2\), \(x-1=1\), so substitution gives

\sum_{n=0}^{\infty}\frac{1}{n+1},

which diverges by the block argument above. Substituting \(x=2\) gives \(1^n=1\), so no alternating signs appear. The interior \((0,2)\) converges by the radius result, and the exterior points do not. Including only the convergent endpoint gives the interval \([0,2)\).

Worked Example: Reversing Which Endpoint Converges

Consider the power series

\sum_{n=0}^{\infty}\frac{(-1)^{n+1}(x-4)^n}{n+1}.

The ratio of successive term magnitudes tends to \(|x-4|\), so the radius is \(1\). The endpoints are \(3\) and \(5\).

At the left endpoint \(x=3\), we have \(x-4=-1\). Thus the \(n\)th term is

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

The endpoint series is the negative of the positive harmonic series, so it diverges. Multiplying all partial sums by \(-1\) does not make an unbounded sequence of positive partial sums converge.

At the right endpoint \(x=5\), \(x-4=1\), and substitution gives

\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n+1}.

This is the negative of the convergent alternating series established above, so it converges. The signs are checked directly: the coefficient contributes \((-1)^{n+1}\), while the endpoint factor \(1^n\) contributes no additional sign. Therefore the interval of convergence is \((3,5]\).

Worked Example: A Shifted and Rescaled Series

Test both endpoints of

\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)3^n}(x-2)^n.

The ratio of successive term magnitudes is

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

It follows that the radius is \(3\), the center is \(2\), and the endpoints are \(-1\) and \(5\).

At \(x=-1\), \(x-2=-3\). Substitution into the \(n\)th term gives

\frac{(-1)^n(-3)^n}{(n+1)3^n} = \frac{1}{n+1}.

The endpoint series is positive and diverges. In particular, the two factors \((-1)^n\) multiply to \(1\); the signs do not alternate.

At \(x=5\), \(x-2=3\), and substitution gives

\frac{(-1)^n3^n}{(n+1)3^n} = \frac{(-1)^n}{n+1}.

This is the convergent alternating series established earlier. The cancellation of \(3^n\) is exact, leaving the alternating signs. The interior \((-1,5)\) converges by the radius result; of the endpoints, only \(5\) converges. The full interval is \((-1,5]\).

A Common Endpoint Mistake

A tempting but incorrect shortcut is to assume that a power series must behave alike at two points because they are the same distance from the center. For the first example, this would incorrectly predict that both endpoints produce the same series. They do not:

\begin{array}{ll} x=0: & \displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1},\\[6pt] x=2: & \displaystyle\sum_{n=0}^{\infty}\frac{1}{n+1}. \end{array}

The correction is to calculate \(x-c\) at each endpoint before simplifying. At the left endpoint it is negative, and raising it to the \(n\)th power may alternate the signs. At the right endpoint it is positive. Equal distances from the center do not mean equal endpoint series.

Check the signs before choosing a test. Write the endpoint value of \(x-c\), raise it to the \(n\)th power, and only then combine it with the coefficient \(a_n\).

Why Independent Endpoint Tests Matter

Finding a radius gives a candidate interval, not a complete answer. The endpoints are the exact places where the radius-based interior and exterior conclusions stop deciding convergence. A complete interval of convergence must record three separate facts: the open interior, the left endpoint result, and the right endpoint result.

A useful record is a small endpoint table:

Point	Substituted series	Result
Left endpoint	Write the series after substituting \(x=c-R\).	Test this series.
Right endpoint	Write the series after substituting \(x=c+R\).	Test this series independently.

This procedure is especially important when coefficients already contain alternating signs. The factor \((-1)^n\) from the left endpoint may preserve those signs, cancel them, or combine with them in another way. The simplified endpoint series, rather than the original appearance of the coefficients, is what should be tested.

Locate the boundary points.
For center \(c\) and radius \(R\), use \(c-R\) on the left and \(c+R\) on the right.

Substitute separately.
Use \(-R\) for \(x-c\) at the left endpoint and \(R\) at the right endpoint.

Simplify each numerical series.
Combine all signs and factors before deciding which convergence test applies.

State the interval with endpoint status.
Include each endpoint exactly when its own substituted series converges.

Check Your Understanding

For each question, distinguish the substitution step from the convergence decision.

For \(\sum_{n=0}^{\infty}a_n(x-c)^n\) with finite radius \(R\), what series results at \(c-R\), and what series results at \(c+R\)?
Why can two endpoints at equal distances from the center yield different convergence behavior?
At \(x=-1\), simplify the \(n\)th term \((-1)^n(-1)^n/(n+1)\). Which familiar positive series results?
For the first worked example, which endpoint converges, and what interval of convergence follows?
What sign calculation should you do before selecting a test for an endpoint series?