Alternating Series

Every test so far has demanded positive terms. But a series whose terms flip sign can converge for a gentler reason: each new term over-corrects the last, so the running total zig-zags inward and homes in on a limit. That is the Alternating Series Test (Leibniz's test).

\sum_{n=1}^{\infty} (-1)^{n+1} b_n = b_1 - b_2 + b_3 - b_4 + \cdots, \qquad b_n > 0.

If the sizes b_n decrease monotonically to zero — that is b_{n+1} \le b_n and b_n \to 0 — the series converges. And you get a remarkable bonus: the error of stopping early is no bigger than the very next term you dropped.

Why the partial sums nest toward a limit

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 proof 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}:

s_{2n+2} - s_{2n} = b_{2n+1} - b_{2n+2} \;\ge\; 0,

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}:

s_{2n+1} - s_{2n-1} = -\,b_{2n} + b_{2n+1} = -(b_{2n} - b_{2n+1}) \;\le\; 0,

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:

s_{2n+1} - s_{2n} = b_{2n+1} \;\ge\; 0 \quad\Longrightarrow\quad s_{2n} \le s_{2n+1}.

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:

L_{\text{odd}} - L_{\text{even}} = \lim_{n\to\infty} (s_{2n+1} - s_{2n}) = \lim_{n\to\infty} b_{2n+1} = 0.

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

\sum_{n=1}^{\infty} (-1)^{n+1} b_n = s, \qquad s_2 \le s_4 \le \cdots \le s \le \cdots \le s_3 \le s_1.

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:

|s - s_n| \;\le\; |s_{n+1} - s_n| = b_{n+1}.

Worked example: the alternating harmonic series

The plain harmonic series diverges, but flip the signs and it springs to life. With b_n = 1/n — positive, decreasing, tending to zero — the test applies, and the value is a celebrated one:

\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \frac12 + \frac13 - \frac14 + \cdots = \ln 2.

The error bound says that summing to the nth term lands within \tfrac{1}{n+1} of \ln 2 — convergence slow enough to be honest, but guaranteed.

Let b_n > 0 with b_{n+1} \le b_n for all n and \displaystyle\lim_{n\to\infty} b_n = 0. Then:

Most convergence tests tell you only that a series converges; the alternating test hands you a free, fully explicit error bar. Because s always lives in the interval [s_n, s_{n+1}] (or [s_{n+1}, s_n]), the moment you stop after term n your worst-case error is exactly the size of the next term:

|s - s_n| \le b_{n+1}.

Want \ln 2 to three decimals? You need b_{n+1} = \tfrac{1}{n+1} < 5 \times 10^{-4}, i.e. about n = 2000 terms — no remainder integral, no tail estimate, just "look at the next term". And the sign of that term even tells you which side of s you are on. It is the most practical error bound in elementary analysis.

Watch the partial sums zig-zag in

The dots are the partial sums of \sum (-1)^{n+1}/n^{p}: each step overshoots the limit, then the next undershoots, the swings shrinking like b_{n+1}. The dashed line is the limit s, and the partial sums stay forever trapped between consecutive ones — the nesting, made visible. A larger p means faster decay and a tighter spiral.