When your calculator works out \ln 2, a sine, or the digits of \pi, it often adds up a series whose terms flip between plus and minus, stopping the moment the next term is too small to matter. This page is about those sign-flipping sums: why they settle down, and how you know exactly when to stop.

Here is a series you already know is doomed. The harmonic series 1 + \tfrac12 + \tfrac13 + \tfrac14 + \cdots piles up positive terms forever and drifts off to infinity — slowly, majestically, but unstoppably. Every test you have met so far would sign its death certificate.

Now do one tiny thing: flip every second sign.

The same numbers, the same sizes, the same lazy 1/n decay — and suddenly the series converges, and to a famous constant no less. Nothing about the sizes of the terms changed; what changed is that each term now partly cancels the one before it. Cancellation is a superpower: a series whose sizes alone would doom it can be rescued by its signs.

Every test so far has demanded positive terms and asked "do the sizes shrink fast enough?" This page's test asks a completely different question: "do the terms take turns pulling in opposite directions?" When they do — and the tugs die away — the running total zig-zags inward and homes in on a limit. That is the Alternating Series Test, also called Leibniz's test after Gottfried Leibniz, who spotted it in the 1670s.

The test

An alternating series strictly alternates sign:

Here the b_n are the sizes — all positive — and the (-1)^{n+1} supplies the sign flips. Leibniz's discovery: if those sizes decrease monotonically to zero, that alone forces convergence. No comparison, no integral, no ratio — just "shrinking tug-of-war".

Read the hypotheses like a pre-flight checklist — there are exactly three boxes to tick:

1. the signs strictly alternate;   2. the sizes decrease, b_{n+1} \le b_n;   3. the sizes tend to zero, b_n \to 0.

Tick all three and you get convergence plus a free, explicit error bar — a luxury almost no other convergence test offers. Miss any one and the test simply doesn't apply (more on that in the traps below).

Watch the partial sums zig-zag in

This is the picture to carry in your head. The bold zig-zag traces the partial sums of \sum (-1)^{n+1}/n^{p}: the first term jumps up to s_1 = 1, the second drags it back down to s_2, the third pushes it up again — but each correction is smaller than the one before, so every swing overshoots the limit by less. The dashed line is the limit s, and the thin converging curves are the \pm b_{n+1} envelope: a shrinking bracket that the partial sums can never escape.

Notice two things as you watch. First, the odd-numbered sums all sit above the dashed line and stair-step down towards it, while the even-numbered sums sit below and climb up — the limit is forever pinched between the last two partial sums, like a ball bouncing inside a closing vice. Second, drag the decay slider: at p = 1 the bracket closes lazily (that's the alternating harmonic series, sauntering towards \ln 2); at p = 3 it snaps shut in a handful of terms. The speed of convergence is exactly the speed at which b_{n+1} dies.

Why the partial sums nest toward a limit

The zig-zag picture practically is the proof — we just have to say it carefully. Assume b_1 \ge b_2 \ge b_3 \ge \cdots \ge 0 with b_n \to 0, and write s_n for the partial sum. The whole argument is watching the even and odd partial sums close in on each other like a vice.

Step 1 — the even partial sums increase. Group the new pair of terms added going from s_{2n} to s_{2n+2}:

which is non-negative because b_{2n+1} \ge b_{2n+2}. So s_2 \le s_4 \le s_6 \le \cdots — the even sums climb.

Step 2 — the odd partial sums decrease. Likewise, going from s_{2n-1} to s_{2n+1}:

so s_1 \ge s_3 \ge s_5 \ge \cdots — the odd sums fall.

Step 3 — every even sum lies below every odd sum. The two run toward each other because a single term separates consecutive partial sums:

So the rising even sums are all bounded above (by s_1) and the falling odd sums are all bounded below (by s_2) — two nested, monotone, bounded sequences. Each converges; call the limits L_{\text{even}} and L_{\text{odd}}.

Step 4 — the two limits coincide. Their gap is a single term, which vanishes:

So L_{\text{even}} = L_{\text{odd}} = s; the even and odd partial sums squeeze onto one common limit, and the series converges:

Step 5 — read off the error bound. Because the limit s is always boxed between consecutive partial sums s_n and s_{n+1}, and those two differ by exactly b_{n+1}, the distance from any partial sum to the true value is at most one term:

Notice where each hypothesis earned its keep: monotone decrease made the even sums climb and the odd sums fall (Steps 1–2), and terms tending to zero welded the two limits together (Step 4). Delete either and the proof — and the theorem — collapses, exactly as the counterexamples in the trap box showed.

Worked examples

Example 1 — the alternating harmonic series passes

Take \sum_{n=1}^{\infty} (-1)^{n+1}/n and run the checklist on b_n = 1/n:

• Positive? Yes: 1/n > 0.
• Decreasing? Yes: \tfrac{1}{n+1} \le \tfrac{1}{n} for every n.
• Tending to zero? Yes: 1/n \to 0.

All three boxes tick, so the series converges — even though its all-positive cousin, the harmonic series, diverges. The value happens to be a celebrity:

(The test itself never tells you what the sum is — the \ln 2 comes from the Taylor series of \ln(1+x) at x = 1. The test's job is only the "converges" verdict, plus the error bar.)

Example 2 — the error bound, used for real

Suppose you want \ln 2 correct to two decimal places — an error below 0.005. The error bound says the error after n terms is at most the next term, b_{n+1} = \tfrac{1}{n+1}, so demand

Two hundred terms. For two decimal places! Check it against reality: s_{10} \approx 0.6456 — off by about 0.048, comfortably inside the promised b_{11} = 1/11 \approx 0.091, but a dreadful approximation. The pattern continues brutally: each extra decimal digit costs roughly ten times as many terms, so six decimals costs about two million terms. The alternating harmonic series converges — but at the speed of continental drift, because its bracket only shrinks like 1/n.

Compare \sum (-1)^{n+1}/n^3: for two-decimal accuracy you need \tfrac{1}{(n+1)^3} \le 0.005, i.e. n + 1 \ge \sqrt[3]{200} \approx 5.85 — just n = 5 terms. Same test, same bracket, wildly different closing speed. You saw exactly this on the slider above.

Example 3 — a series the test rejects

Does \sum_{n=1}^{\infty} (-1)^{n+1} \dfrac{n}{2n+1} converge? It alternates, and the sizes b_n = \dfrac{n}{2n+1} even decrease… towards \tfrac12, not 0. Hypothesis 3 fails, so Leibniz's test is silent — but the divergence test is not: terms of size near \tfrac12 forever mean the partial sums bounce between two values about \tfrac12 apart and never settle. Divergent. Moral: check b_n \to 0 first — it is the quickest box to tick and the deadliest to skip.

See it explained