Alternating series converge for a very visual reason: the partial sums zigzag, and the zigzags shrink. You used the test in Calculus 2; here we prove it with nested intervals and the Cauchy criterion, and get a sharp error bound for free.
🔑 Key idea
Alternating series test. If $b_k$ decreases and $b_k \to 0$, then $S = \sum_{k=1}^{\infty} (-1)^{k+1} b_k$ converges, $S$ lies between any two consecutive partial sums, and $|S - s_n| < b_{n+1}$ (when $b_k$ strictly decreases).
💡 Why it works
The even partial sums increase: $s_{2m+2} = s_{2m} + (b_{2m+1} - b_{2m+2}) \ge s_{2m}$. The odd partial sums decrease: $s_{2m+1} = s_{2m-1} - (b_{2m} - b_{2m+1}) \le s_{2m-1}$. And every even sum is below every odd sum. So the even sums increase and are bounded above, hence converge, and the odd sums decrease to a limit too; their difference $s_{2m+1} - s_{2m} = b_{2m+1} \to 0$, so the limits agree. Call it $S$. Since $S$ sits between $s_n$ and $s_{n+1}$, its distance from $s_n$ is less than $|s_{n+1} - s_n| = b_{n+1}$. $\blacksquare$
The alternating harmonic series on a number line: each step goes the other way and is shorter, so the sum is trapped between consecutive partial sums.
The error bound is one of the few in analysis that you get completely free: no derivatives to estimate, just the next term. It also tells you the sign of the error. If the last term you added was positive, you have overshot, and $s_n > S$; if it was negative, $s_n < S$. Averaging two consecutive partial sums is often much closer to $S$ than either one, because the true sum sits between them.
Worked examples
✏️ Worked example: How many terms are enough?
How many terms of $\sum (-1)^{k+1}\frac{1}{k^{2}}$ guarantee an error below $0.001$?
The error after $n$ terms is less than $b_{n+1} = \frac{1}{(n + 1)^{2}}$.
$\frac{1}{(n + 1)^{2}} < 0.001$ means $(n + 1)^{2} > 1000$, so $n + 1 \ge 32$.
Answer: $n = 31$ terms (or more).
✏️ Worked example: Above or below?
For $\sum (-1)^{k+1}\frac{1}{k}$, is $s_5$ above or below the sum?
The fifth term is $+\frac{1}{5}$: adding it overshoots.
The error bound uses the next term. After $n$ terms, the bound is $b_{n+1}$, the first term left out, not $b_n$. Also check the hypotheses: if $b_k$ is not decreasing (say $\frac{1}{k}$ for odd $k$ and $\frac{1}{k^{2}}$ for even $k$), the series can diverge even though $b_k \to 0$ and the signs alternate.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.