AP Calculus BC

When the Alternating Series Test Does Not Apply

The Alternating Series Test needs two conditions on the term sizes: they must decrease, and they must tend to zero. Alternating signs on their own prove nothing. If the sizes tend to a nonzero value the series diverges outright, whatever the pattern of signs.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

1. 1

   The term sizes tend to zero

   $$\lim_{n \to \infty} b_n = 0$$

   This is the condition that makes the partial sums close in on a value: consecutive partial sums differ by exactly $b_n$, so if the sizes do not shrink to nothing, the sums keep jumping by a fixed amount and can never settle.

   Drop it and the theorem fails

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

   Alternating terms of size n over n plus one

   The sizes $\frac{n}{n+1}$ do increase, and they head toward 1 rather than 0. The check confirms they are still about $0.999999$ at the millionth term. The signs alternate perfectly and the series diverges anyway, by the nth term test, because the terms themselves never approach zero.

2. 2

   The term sizes are decreasing

   $$b_{n+1} \le b_n \text{ for all large } n$$

   Decreasing sizes are what make the partial sums nest inside one another, each one trapped between the previous two. Without that nesting the sums can wander even when the sizes tend to zero. Building an example takes a term defined by cases, one rule for odd indices and another for even ones, which no single elementary formula produces, so none is displayed here. The condition is genuinely needed: it is not an artefact of the proof.

   No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

The counterexample above is checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

1. Group the partial sums by parity. The even-indexed ones increase, because each step adds $b_{2k+1} - b_{2k+2} \ge 0$, using the decreasing hypothesis.
2. The odd-indexed ones decrease, by the same pairing shifted along.
3. Every even partial sum stays below every odd one, so both sequences are monotone and bounded, and both converge.
4. The gap between them is $b_n$, which tends to 0 by the second hypothesis, so the two limits coincide and the series converges. Each hypothesis powers a different half: decreasing gives the monotonicity, vanishing closes the gap.

What it does not say