Endpoint Analysis

Why the Boundary Needs Its Own Analysis

The Root Test for Power Series determines what happens strictly inside or strictly outside the radius of convergence. At the boundary, however, the root calculation reaches its threshold and does not settle convergence. Endpoint Analysis begins by writing down the two series at those boundary points and testing them directly.

Suppose a power series centered at \(a\) has finite, positive radius of convergence \(R\). Its boundary points are \(a+R\) and \(a-R\). At the first point, the powers \((x-a)^n\) become \(R^n\); at the second, they become \((-R)^n\). Thus the two endpoint series differ by a factor of \((-1)^n\) on each term. That sign change can alter convergence, but it does not guarantee that one endpoint converges whenever the other does.

Definition (Endpoint Series): For a power series \(\sum_{n=0}^{\infty}c_n(x-a)^n\) with finite positive radius \(R\), the endpoint series are the series obtained by substituting \(x=a+R\) and \(x=a-R\). If \(b_n=c_nR^n\), they are \(\sum_{n=0}^{\infty}b_n\) and \(\sum_{n=0}^{\infty}(-1)^n b_n\), respectively.

This notation isolates the issue: the endpoint test concerns the terms \(b_n\) and the alternating sign pattern, not the location of the center. The convergence-inside-and-outside theorem and the Root Criterion already establish what happens away from these endpoints. At \(x=a\), the series converges to \(c_0\), as established by the Value at the Center proposition. The remaining task is to classify the two boundary series individually.

A Parity Decomposition of the Endpoints

The right endpoint adds all the \(b_n\), while the left endpoint adds even-indexed terms and subtracts odd-indexed terms. It is useful to separate those two groups. Write \(P_N\) for the \(N\)th partial sum at the right endpoint and \(Q_N\) for the \(N\)th partial sum at the left endpoint. Also let \(E_N\) be the sum of the even-indexed \(b_n\) with \(n\leq N\), and \(O_N\) the sum of the odd-indexed \(b_n\) with \(n\leq N\). Then

P_N=E_N+O_N,\qquad Q_N=E_N-O_N.

These finite identities suggest that convergence at both endpoints is equivalent to convergence of the even and odd subseries separately. The next result makes that precise. A subseries here means the series formed by retaining the indicated terms in their original order.

Theorem (Parity Decomposition at the Endpoints): The endpoint series \(\sum_{n=0}^{\infty}b_n\) and \(\sum_{n=0}^{\infty}(-1)^n b_n\) both converge if and only if the even-indexed subseries \(\sum_{k=0}^{\infty}b_{2k}\) and the odd-indexed subseries \(\sum_{k=0}^{\infty}b_{2k+1}\) both converge.

Proof. For each \(N\), the finite sums satisfy

E_N=\frac{P_N+Q_N}{2},\qquad O_N=\frac{P_N-Q_N}{2}.

If both endpoint series converge, then \(P_N\) and \(Q_N\) have finite limits. These identities show that \(E_N\) and \(O_N\) also have finite limits. As \(N\) increases, \(E_N\) is the sequence of partial sums of the even-indexed subseries, with repeated values when \(N\) is odd; thus that subseries converges. The same reasoning applies to the odd-indexed subseries.

Conversely, suppose the even- and odd-indexed subseries converge, with sums \(E\) and \(O\). Then \(E_N\to E\) and \(O_N\to O\). The finite identities \(P_N=E_N+O_N\) and \(Q_N=E_N-O_N\) give \(P_N\to E+O\) and \(Q_N\to E-O\). Hence both endpoint series converge. This proves both directions. \(\square\)

A useful consequence is that when both endpoint sums exist, their sum and difference recover the parity-subseries sums. This can help analyze a power series whose coefficients have a natural pattern by parity.

Corollary (Endpoint Sums Recover Parity Sums): If the endpoint series converge to \(S_+\) and \(S_-\), respectively, and the even- and odd-indexed subseries converge to \(E\) and \(O\), then

E=\frac{S_++S_-}{2},\qquad O=\frac{S_+-S_-}{2}.

Proof. The theorem gives \(S_+=E+O\) and \(S_-=E-O\). Adding these equations yields \(S_++S_-=2E\), and subtracting the second from the first yields \(S_+-S_-=2O\). Dividing by \(2\) gives the stated identities. \(\square\)

Worked Examples: Classifying Boundary Behavior

Worked Example: One Endpoint Converges and the Other Diverges

Consider the power series centered at zero

\sum_{n=1}^{\infty}\frac{x^n}{n}.

For \(n\geq1\), the coefficient-root values satisfy

\left|\frac{1}{n}\right|^{1/n}=n^{-1/n}\longrightarrow1.

The Cauchy–Hadamard Formula therefore gives radius \(R=1\). At the right endpoint \(x=1\), the series is \(\sum_{n=1}^{\infty}1/n\), which diverges by the \(p\)-Series Convergence Criterion with \(p=1\). At the left endpoint \(x=-1\), it is \(\sum_{n=1}^{\infty}(-1)^n/n\). The magnitudes \(1/n\) decrease to zero, so the Alternating Series Test gives convergence. It is not absolutely convergent, since the absolute-value series is the divergent harmonic series. Thus this power series converges conditionally at one endpoint and diverges at the other.

Worked Example: Both Endpoints Diverge Even Though the Terms Vanish

Define coefficients by \(c_{2k}=1/\sqrt{2k}\) for \(k\geq1\), and \(c_{2k+1}=0\) for \(k\geq0\). Consider the power series

\sum_{n=1}^{\infty}c_nx^n = \sum_{k=1}^{\infty}\frac{x^{2k}}{\sqrt{2k}}.

The coefficient roots along even indices satisfy

|c_{2k}|^{1/(2k)} = \left(\frac{1}{\sqrt{2k}}\right)^{1/(2k)} \longrightarrow1.

At odd indices the coefficient roots are zero. Since the even-indexed roots tend to \(1\), the limsup of all coefficient roots is \(1\); the radius is \(R=1\). At either endpoint, \(x=1\) or \(x=-1\), the terms with odd indices vanish and \(x^{2k}=1\). Each endpoint series is therefore

\sum_{k=1}^{\infty}\frac{1}{\sqrt{2k}} = \frac{1}{\sqrt{2}}\sum_{k=1}^{\infty}\frac{1}{\sqrt{k}}.

The \(p\)-Series Convergence Criterion with \(p=1/2\) shows that this diverges. Nevertheless, the individual terms of the power series at either endpoint tend to zero: along even indices their magnitudes are \(1/\sqrt{2k}\to0\), and along odd indices they are zero. This illustrates why the Necessary Condition for Series Convergence is only a necessary condition. Terms tending to zero does not suffice for convergence.

Worked Example: Both Endpoints Converge Conditionally

For \(k\geq0\), define coefficients by

c_{2k}=\frac{(-1)^k}{k+1}, \qquad c_{2k+1}=\frac{(-1)^k}{k+1}.

The coefficient roots have limsup \(1\). Indeed, at even indices their magnitudes have roots \((k+1)^{-1/(2k)}\to1\), and at odd indices their roots are \((k+1)^{-1/(2k+1)}\to1\). Thus the radius is \(R=1\). At \(x=1\), the even- and odd-indexed subseries are each

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

Each converges by the Alternating Series Test: its positive magnitudes decrease to zero. The Parity Decomposition Theorem now gives convergence at \(x=1\). At \(x=-1\), the even-indexed terms are unchanged and the odd-indexed terms change sign. The even subseries still converges, and the odd subseries is the negative of a convergent alternating series. The theorem therefore gives convergence at \(x=-1\) as well.

Neither endpoint convergence is absolute. The sum of the absolute values at either endpoint is

\sum_{k=0}^{\infty}\left(\frac{1}{k+1}+\frac{1}{k+1}\right) = 2\sum_{k=0}^{\infty}\frac{1}{k+1},

which diverges by the harmonic-series case of the \(p\)-Series Convergence Criterion. Hence both endpoint series converge conditionally. The example also shows why endpoint behavior cannot be summarized by a rule that says one endpoint must converge while the other diverges.

A Reliable Endpoint Procedure

For a power series with \(0<R<\infty\), a careful classification keeps the strict interior conclusions separate from the boundary analysis. The following sequence of steps prevents the Root Test from being used beyond what it establishes.

Identify the two boundary inputs.
They are \(a+R\) and \(a-R\). The center and the points strictly inside or outside the radius are handled by earlier radius results.

Substitute at each endpoint.
Write the resulting numerical series explicitly. In particular, define \(b_n=c_nR^n\), so the two series are \(\sum b_n\) and \(\sum(-1)^n b_n\).

Test each series on its own terms.
Check absolute convergence first when appropriate; otherwise consider tests suited to the signs and magnitudes, such as comparison or the Alternating Series Test.

Use parity when the coefficients have structure.
If both endpoint series are under consideration, the Parity Decomposition Theorem reduces their joint convergence to convergence of the even- and odd-indexed subseries.

There is an important distinction between a fact that holds at both endpoints and a fact that decides their convergence. The Absolute Endpoint Convergence Is Symmetric theorem established earlier says that absolute convergence at one endpoint implies absolute convergence at the other. But conditional convergence need not be symmetric, as the first worked example shows. Nor does convergence at one endpoint in general settle the other endpoint.

Another common pitfall is to treat a root-test equality as an endpoint classification. When the endpoint is exactly distance \(R\) from the center, the root criterion reaches its boundary case and is inconclusive. The endpoint series may converge absolutely, converge conditionally, or diverge. The coefficient pattern and signs must be examined directly. Even when the terms tend to zero, as in the second example, their series may still diverge.

Takeaway: For finite positive radius \(R\), write the endpoint terms as \(b_n=c_nR^n\). The two boundary series are \(\sum b_n\) and \(\sum(-1)^n b_n\). They both converge exactly when the even- and odd-indexed subseries both converge, but either endpoint may behave differently from the other.

Check Your Understanding

Use endpoint substitution and the parity decomposition to answer the following questions.

For a power series with center \(a\) and radius \(R\), what numerical series results at \(a+R\), and what series results at \(a-R\), when \(b_n=c_nR^n\)?
If both endpoint series converge, what does the Parity Decomposition Theorem imply about the even- and odd-indexed subseries?
If the endpoint sums are \(S_+\) and \(S_-\), express the sum of the even-indexed subseries in terms of \(S_+\) and \(S_-\).
Why does convergence of the terms to zero fail to establish convergence of an endpoint series?
Can one endpoint converge conditionally while the other diverges? Explain what feature of the endpoint series can cause this difference.

What You'll Learn