The Alternating Series Test

Over, then under

In the alternating harmonic series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots$1 − 1/2 + 1/3 − 1/4 + 1/5 − …, the terms are the harmonic series's terms with every second one subtracted.

Its partial sums are 1, then 0.5, then about 0.833, 0.583, 0.783, 0.617. Each added term carries the total up past where it is heading, and each subtracted term brings it back down below, by a little less each time. The totals close in on one number from both sides. That number is ln 2, about 0.693.

The partial sums 1, 0.5, 0.833, 0.583, 0.783: each hop is a term, $-\frac{1}{2}$−1/2, $+\frac{1}{3}$+1/3, $-\frac{1}{4}$−1/4, $+\frac{1}{5}$+1/5, and each crosses the gold mark at 0.693.

The three conditions

Write the series as $b_1-b_2+b_3-b_4+\dots$b₁ − b₂ + b₃ − b₄ + …, with every $b_n$bₙ positive. The alternating series test says it converges if three things hold: the signs alternate; the sizes decrease, $b_1\ge b_2\ge b_3\ge \dots$b₁ ≥ b₂ ≥ b₃ ≥ …; and the sizes tend to 0.

For $1-\frac{1}{2}+\frac{1}{3}-\dots$1 − 1/2 + 1/3 − …, the sizes are $\frac{1}{n}$1/n: they fall, since $\frac{1}{n+1}<\frac{1}{n}$1/(n + 1) < 1/n, and they tend to 0. The series converges.

All three are needed. In $\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\dots$1/2 − 2/3 + 3/4 − 4/5 + …, the sizes $\frac{n}{n+1}$n/(n + 1) tend to 1, not 0, and by the nth term test the series diverges. And sizes that tend to 0 without falling steadily can fail as well, as in a series further down.

Why the limit is trapped

Look at the odd partial sums $S_1$S₁, $S_3$S₃, $S_5$S₅, …. From $S_1$S₁ to $S_3$S₃ the series subtracts $b_2$b₂ and adds $b_3$b₃, and since $b_3\le b_2$b₃ ≤ b₂ the total goes down, or stays put. So the odd sums never rise: 1, 0.833, 0.783, …. In the same way, from $S_2$S₂ to $S_4$S₄ the series adds $b_3$b₃ and subtracts $b_4$b₄, so the even sums never fall: 0.5, 0.583, 0.617, ….

Every even sum is below every odd sum, and the gap between $S_n$Sₙ and $S_{n+1}$Sₙ₊₁ is $b_{n+1}$bₙ₊₁, which tends to 0. The falling odd sums and the rising even sums close in on the same number, S, and it lies between any two consecutive partial sums.

The partial sums $S_1$S₁ to $S_{10}$S₁₀ of $1-\frac{1}{2}+\frac{1}{3}-\dots$1 − 1/2 + 1/3 − …, with the vertical scale stretched. The odd ones fall from 1 to about 0.746; the even ones rise from 0.5 to about 0.646; the gold line at ln 2, about 0.693, runs between them.

The error from stopping early

Since S lies between $S_n$Sₙ and $S_{n+1}$Sₙ₊₁, and those two are $b_{n+1}$bₙ₊₁ apart, stopping after n terms leaves an error of at most $b_{n+1}$bₙ₊₁: $|S-S_n|\le b_{n+1}$|S − Sₙ| ≤ bₙ₊₁, the size of the first term left out.

After four terms, $S_4=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}=\frac{7}{12}$S₄ = 1 − 1/2 + 1/3 − 1/4 = 7/12, about 0.583, and the first term left out is $\frac{1}{5}$1/5. So the sum is within 0.2 of 0.583, and it lies between $S_4\approx 0.583$S₄ ≈ 0.583 and $S_5\approx 0.783$S₅ ≈ 0.783. The actual error is 0.693 − 0.583, about 0.110, inside the bound.

To be sure of an error of at most 0.01, the first term left out must be at most 0.01: $\frac{1}{n+1}\le 0.01$1/(n + 1) ≤ 0.01, so $n+1\ge 100$n + 1 ≥ 100, and 99 terms are enough.

A series whose sizes fall faster closes in faster. In $1-\frac{1}{4}+\frac{1}{9}-\frac{1}{16}+\frac{1}{25}-\dots$1 − 1/4 + 1/9 − 1/16 + 1/25 − …, the sizes are $\frac{1}{n^2}$1/n². After five terms the partial sum is about 0.8386, and the bound is $\frac{1}{36}$1/36, about 0.0278. The sum is $\frac{{\pi }^2}{12}$π²/12, about 0.8225, so the actual error is about 0.0161.

When the sizes do not fall

Take $2-1+1-\frac{1}{2}+\frac{2}{3}-\frac{1}{3}+\frac{2}{4}-\frac{1}{4}+\dots$2 − 1 + 1 − 1/2 + 2/3 − 1/3 + 2/4 − 1/4 + …, made of pairs $\frac{2}{k}-\frac{1}{k}$2/k − 1/k for k = 1, 2, 3, …. The signs alternate and the sizes tend to 0, but they do not fall steadily: after $\frac{1}{3}$1/3 comes $\frac{2}{4}$2/4, which is larger.

Each pair adds to $\frac{1}{k}$1/k, so after 2n terms the partial sum is $1+\frac{1}{2}+\cdots +\frac{1}{n}$1 + 1/2 + … + 1/n, a partial sum of the harmonic series. The series diverges. The decreasing condition is not a technicality.

Absolute and conditional convergence

Drop the signs of $1-\frac{1}{2}+\frac{1}{3}-\dots$1 − 1/2 + 1/3 − … and what is left is the harmonic series, which diverges. The series converges only because of its signs: it is conditionally convergent.

Drop the signs of $1-\frac{1}{4}+\frac{1}{9}-\dots$1 − 1/4 + 1/9 − … and what is left is $\sum \frac{1}{n^2}$Σ 1/n², which converges. A series that still converges with every term made positive is absolutely convergent; the signs are not needed.

The usual mistakes

Giving the last term used as the error bound. After four terms the bound is $\frac{1}{5}$1/5, the first term left out, not $\frac{1}{4}$1/4.

Checking only that the signs alternate and the terms tend to 0. The sizes must also decrease.

Keeping an old bound after adding terms. The bound moves with n: $\frac{1}{5}$1/5 after four terms, $\frac{1}{100}$1/100 after 99.

Calling $1-\frac{1}{2}+\frac{1}{3}-\dots$1 − 1/2 + 1/3 − … absolutely convergent. Without its signs it is the harmonic series.

A robot arm's corrections

In the application below, a robot arm moves $\frac{12}{n}$12/n millimeters on step n, in alternate directions. The test shows the arm settles, and the first move left out bounds how far it still has to go.