Writing Alternating-Series Convergence and Error Claims

Why Alternating-Series Justifications Need Two Checks

An alternating series changes sign from one term to the next. This sign pattern can make a series converge even when the corresponding positive-term series does not. To justify convergence, however, it is not enough to point out that the signs alternate. The magnitudes must also satisfy specific conditions.

Those same conditions support an important estimate: after adding a finite number of terms, the error is no larger than the magnitude of the first term left out. A careful written justification should identify the term magnitudes, verify the conditions, state the conclusion, and use the error bound with the correct index.

The Alternating-Series Test

A series is in alternating form when its terms alternate in sign and can be written using positive magnitudes \(b_n\). A common representation is \(\sum_{n=1}^{\infty}(-1)^{n+1}b_n\), where \(b_n\ge 0\). The first term is positive in this convention; changing the initial sign does not change the convergence test.

“Nonincreasing” means \(b_{n+1}\le b_n\) for every relevant \(n\). Some courses state the test using “decreasing”; strict decrease is sufficient, but it is not necessary. The condition may hold only eventually: a finite number of initial terms do not affect whether the infinite series converges. For the error bound, choose \(N\) so that the terms beginning with \(b_{N+1}\) lie in the nonincreasing part of the sequence.

The two checks have distinct jobs. Nonincreasing magnitudes ensure that successive terms do not grow, while the limit of zero ensures that the terms eventually become arbitrarily small. Alternating signs alone establish neither condition.

Why the Test and Error Bound Work

First consider the case in which \(b_n\) is nonincreasing from the beginning. Let \(S_N\) denote the partial sums. The even-indexed partial sums are nondecreasing, because

The odd-indexed partial sums are nonincreasing, because

For each \(m\), the even partial sum is no greater than the next odd partial sum:

Also, the difference between these neighboring partial sums tends to zero because \(b_{2m+1}\to0\). The even partial sums are bounded above by the odd partial sums, and the odd partial sums are bounded below by the even partial sums. Each subsequence therefore has a limit, and the difference between their limits is zero. Thus both subsequences have the same limit. Since every partial sum belongs to one of these two subsequences, the full sequence of partial sums converges.

The same reasoning explains the remainder estimate. After \(S_N\), the tail begins with \(b_{N+1}\), up to an overall sign. Consider the tail with its first sign made positive:

Pairing consecutive terms gives nonnegative differences:

The tail is also no greater than \(b_{N+1}\): it starts at \(b_{N+1}\) and subtracts a sum of nonnegative paired differences, with the same conclusion obtained by taking limits of finite partial tails. Therefore its magnitude is between zero and \(b_{N+1}\). Restoring the possible overall sign does not change its magnitude, so \(|S-S_N|\le b_{N+1}\). If the conditions begin only after an initial index, apply this argument to the tail that starts in that range; adding or removing finitely many terms does not affect convergence.

Worked Examples

Writing a Complete Justification

In a free-response solution, the conclusion should follow from explicit checks rather than from the phrase “it alternates.” A concise justification generally makes these points in order:

For instance, a complete conclusion might say: “The magnitudes are positive, decrease for \(n\ge2\), and approach zero. Therefore the alternating series converges by the alternating-series test. After \(N\ge2\) terms, its error is at most \(b_{N+1}\).” The inequality or limit calculation supporting each claim should appear before the conclusion.

Common Pitfalls

A frequent error is to treat the remainder bound as an exact error or as a test that works in both directions. In the \(1/n^2\) example, after 30 terms the calculation \(\frac{1}{31^2}>0.001\) does not establish \(|S-S_{30}|>0.001\). It establishes only that the alternating-series bound is too large to guarantee an error of \(0.001\) or less. The actual error may still be below that tolerance.

Another incomplete argument is, “The terms alternate, so the series converges.” Alternation alone is insufficient. For example, the magnitudes \(b_n=1+\frac1n\) tend to \(1\), not zero. The term of the alternating series \((-1)^{n+1}b_n\) therefore does not tend to zero, so the series cannot converge. Always check both eventual nonincrease and the zero limit.

Finally, keep the index straight. If the partial sum includes terms through \(N\), the first omitted magnitude is \(b_{N+1}\), not \(b_N\). The estimate concerns the tail after the included terms.

Why These Claims Matter

Alternating-series problems often ask for more than a convergence label. A complete answer may need to justify the test's hypotheses, estimate how accurate a partial sum is, or determine how many terms guarantee a specified precision. Separating these tasks makes the reasoning reliable: the monotonicity and limit establish convergence, while the next omitted term supplies a one-sided guarantee on the size of the remainder.

The key wording is precise: “The error is at most the first omitted magnitude” is a theorem-backed guarantee. “The error equals the first omitted magnitude,” or “a failed bound proves the error is too large,” is not. State exactly what the test establishes, and do not infer more than its inequality allows.