Alternating-Series Remainders at a Boundary

Why Remainder Bounds Matter at an Endpoint

A power series can behave differently at its boundary than it does inside its interval of convergence. Once a boundary point is substituted, the question becomes one about an ordinary numerical series. If the terms alternate in sign, verifying the size and behavior of their magnitudes can do more than establish convergence: it can also quantify how accurately a finite partial sum approximates the sum.

The key estimate is especially practical. If the magnitudes decrease to zero, then after adding \(N\) terms, the error is no larger than the magnitude of the next term. The important work is checking the hypotheses for the actual endpoint terms; alternation by itself is not enough.

Check the Endpoint Terms, Not Just the Coefficients

Suppose a power series is written as \(\sum_{n=1}^{\infty}c_n(x-x_0)^n\). At a particular endpoint \(x=b\), its terms become \(c_n(b-x_0)^n\). To use an alternating-series remainder bound, these endpoint terms must be expressible as \((-1)^{n-1}a_n\), with \(a_n\geq 0\). Then check both required properties: \(a_{n+1}\leq a_n\) (at least from some index onward) and \(\lim_{n\to\infty}a_n=0\).

The factor \((b-x_0)^n\) may create the alternating signs, or the coefficients may already alternate. Either way, inspect the terms after substituting the endpoint. A coefficient sequence decreasing to zero does not, on its own, establish the same facts for the endpoint terms if other factors are present.

If the conditions hold only after some index, the finite initial segment does not affect whether the series converges. The same estimate applies to partial sums sufficiently far along, once the decreasing condition covers the terms beginning at \(N+1\).

Why the Alternating-Series Estimate Works

First consider a sequence whose magnitudes are nonincreasing from the beginning. Separate the partial sums into even and odd indices. Each new pair of terms increases an even partial sum, while each new pair decreases an odd partial sum. Those two families of partial sums therefore move toward one another.

The even partial sums are increasing and bounded above, so they have a limit. The odd partial sums are decreasing and bounded below, so they also have a limit. Since their difference tends to zero, those limits are equal. Thus the full sequence of partial sums converges to a common value \(S\).

For the error estimate, consecutive partial sums differ by one term: \(S_{N+1}-S_N=(-1)^N a_{N+1}\). The limit \(S\) lies between \(S_N\) and \(S_{N+1}\): when \(N\) is even, the even partial sum is a lower bound and the next odd partial sum is an upper bound; when \(N\) is odd, their roles reverse. Therefore the distance from \(S_N\) to \(S\) cannot exceed the gap between these consecutive partial sums, which is \(a_{N+1}\). The error points in the direction of the next term. This proves both the remainder bound and its sign.

In particular, the sum lies between the \(N\)-term partial sum and the partial sum obtained by adding the next term. These bounds can give a useful interval for the sum without requiring its exact value.

Worked Endpoint Examples

A Common Mistake: Alternation Alone Is Not Enough

A tempting but incorrect shortcut is to see alternating signs and immediately use the remainder theorem. For example, consider \(\sum_{n=1}^{\infty}(-1)^{n-1}n/(n+1)\). Its magnitudes \(a_n=n/(n+1)\) increase toward \(1\), rather than decrease to zero. In particular, the terms do not tend to zero: \(a_n\to1\).

The term test says that a convergent series must have terms tending to zero. Here that necessary condition fails, so the series diverges. It would be wrong to claim that the error after \(N\) terms is at most \((N+1)/(N+2)\) by the alternating-series remainder theorem: its required hypotheses are not satisfied. A remainder estimate is justified by verified conditions, not by the visual pattern of plus and minus signs.

Using the Bound to Choose a Partial Sum

The estimate also provides a simple way to decide how many terms are needed for a specified accuracy. If \(a_n\) is nonincreasing and tends to zero, then the error after \(N\) terms is at most \(a_{N+1}\). Thus it is enough to choose \(N\) so that \(a_{N+1}\leq\varepsilon\), where \(\varepsilon>0\) is the desired error tolerance. This gives a guaranteed bound; it does not claim that the actual error equals the first omitted term.

At a power-series boundary, the steps are therefore specific: substitute the endpoint, write the resulting terms as alternating signs times nonnegative magnitudes, verify monotonicity and the zero limit, and only then apply the bound. The endpoint substitution matters because the remainder concerns the series at that point, not the power series in the abstract.