In an alternating series the signs take turns: $b_1 - b_2 + b_3 - b_4 + \cdots$ with every $b_n > 0$. The canceling helps a lot: such a series can converge even when $\sum b_n$ diverges. The test uses the The Divergence Test idea and monotone sequences.
Why it works
Suppose $b_n$ decreases to 0. Look at the even partial sums: $S_{2k+2} = S_{2k} + (b_{2k+1} - b_{2k+2}) \ge S_{2k}$, so they increase. The odd ones decrease: $$S_{2k+3} = S_{2k+1} - (b_{2k+2} - b_{2k+3}) \le S_{2k+1}.$$ And every even sum is below every odd sum. So the even sums rise to some limit, the odd sums fall to some limit, and the gap between them, a single term $b_{2k+1}$, goes to 0. Both limits are the same number $S$. Because $S$ always lies between two consecutive partial sums, $|S - S_N| \le b_{N+1}$.
Key idea
Alternating series test: if $b_n > 0$, $b_{n+1} \le b_n$ (eventually), and $b_n \to 0$, then $\sum(-1)^{n+1}b_n$ converges. Error bound: $|S - S_N| \le b_{N+1}$, the first term left out.
Worked example: The alternating harmonic series
Does $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$ converge?
- $b_n = \frac{1}{n}$ is positive and decreasing, and $\frac{1}{n} \to 0$.
- So it converges by the alternating series test (its sum is $\ln 2$), even though $\sum\frac{1}{n}$ diverges.
Answer: It converges.
Worked example: Terms that do not shrink
Does $\sum_{n=1}^{\infty}(-1)^{n}\frac{n}{n + 1}$ converge?
- The sizes $\frac{n}{n + 1} \to 1$, so the terms do not go to 0.
- It diverges by the divergence test. Alternating signs cannot rescue terms that stay large.
Answer: It diverges.
Worked example: An error bound
Estimate $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}$ with $S_5$, and bound the error.
- $S_5 = 1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \frac{1}{25} \approx 0.8386$.
- The error is at most $b_6 = \frac{1}{36} \approx 0.028$.
- So the sum is between $0.8386 - 0.028$ and $0.8386 + 0.028$. (It is $\frac{\pi^{2}}{12} \approx 0.8225$.)
Answer: $S_5 \approx 0.8386$, error at most $\frac{1}{36}$
Common mistake
Skipping the check that the terms decrease. Take $1 - \frac{1}{4} + \frac{1}{3} - \frac{1}{16} + \frac{1}{5} - \frac{1}{36} + \cdots$ (odd terms $\frac{1}{n}$, even terms $\frac{1}{n^{2}}$). The signs alternate and the terms go to 0, but they do not decrease, and the series diverges: the positive terms alone add like a harmonic series. Check that $b_{n+1} \le b_n$, for instance by showing $f'(x) < 0$ for the function that gives $b_n$.