The Alternating Series Test
When a series flips sign every term, a new kind of convergence kicks in: if the term sizes shrink to 0, the partial sums zig-zag inward and squeeze onto a single value — and the error from stopping early is never bigger than the first term you leave out.
By the end you'll be able to apply the Alternating Series Test to decide convergence, and use the remainder bound \(|R_N| \le b_{N+1}\) to say exactly how many terms you need for a given accuracy.
Predict: stop the alternating harmonic series at k = 5 terms. Is the true sum within one more term (\(b_6\)) of where you stopped? Step k to check — then keep stepping and watch the error bracket shrink.
Each dot is a partial sum \(S_k\) — the same running totals you met with series basics, except here they oscillate instead of climbing: each new term overshoots the true sum in the opposite direction, by a shrinking amount. The bracket at the last dot has half-width \(b_{k+1}\), the first omitted term. Hover or focus any dot for its exact value.
S₅ = 0.7833 · bound b₆ = 0.1667 · actual gap |S − S₅| = 0.0902 — inside the bound ✓
An alternating series \(\sum (-1)^{n+1} b_n\) with \(b_n \ge 0\) converges whenever the \(b_n\) are eventually decreasing and tend to 0 — and stopping after \(N\) terms leaves an error of at most \(b_{N+1}\), the first term you omit.
What the test says
An alternating series flips sign every step: \[ \sum_{n=1}^{\infty} (-1)^{n+1} b_n = b_1 - b_2 + b_3 - b_4 + \cdots \qquad (b_n \ge 0), \] or with \((-1)^n\) out front instead. The tests you've used so far — divergence, integral, comparison — assume positive terms, so sign alternation needs its own criterion. Here it is: if \(\{b_n\}\) is eventually decreasing and \(\lim_{n\to\infty} b_n = 0\), then \(\sum (-1)^{n+1} b_n\) converges. Both hypotheses matter: if \(b_n \not\to 0\), the test says nothing — and the divergence test already shows the series diverges. And always factor out the sign first: \(b_n\) is the size of the term, so check decrease and limit-to-zero on \(b_n \ge 0\), not on the signed terms.
Picture the partial sums in the demo above: each new term overshoots the limit in the opposite direction from the last, but by a shrinking amount, so the sums zig-zag inward and squeeze onto a single value \(S\). That picture is also the proof of the error bound: \(S\) is always trapped between consecutive partial sums — between \(S_N\) and \(S_{N+1}\) for every \(N\) — and those two differ by exactly \(b_{N+1}\), so \(|S - S_N| \le b_{N+1}\).
That trapped-between picture is what makes alternating series a practical approximation tool. To hit a desired tolerance, you don't need to estimate the whole tail — just find how many terms make the next term smaller than the tolerance and stop there. That's the Alternating Series Estimation Theorem: if the series satisfies the test, then \[ |R_N| = |S - S_N| \le b_{N+1}. \] It's far more convenient than the general tail analysis needed for positive series — one term size tells you your worst-case error.
Conditional convergence
The alternating harmonic series \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\) is the classic example: \(b_n = \frac{1}{n}\) is decreasing and tends to 0, so the series converges — even though the corresponding positive series \(\sum \frac{1}{n}\) (the harmonic series) diverges. A series like this is called conditionally convergent: convergent as signed, divergent in absolute value. So the test only proves convergence — never absolute convergence; that distinction is where the ratio and root tests come in next.
Approximate \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!}\) (which equals \(1 - 1/e\)) to within \(0.001\). Here \(b_n = 1/n!\): \(b_4 = 1/24 \approx 0.0417\), \(b_5 = 1/120 \approx 0.00833\), \(b_6 = 1/720 \approx 0.00139\), \(b_7 = 1/5040 \approx 0.000198 < 0.001\). So summing through \(n = 6\) — \(S_6 = 1 - \tfrac12 + \tfrac16 - \tfrac1{24} + \tfrac1{120} - \tfrac1{720}\) — and stopping there guarantees error at most \(b_7 < 0.001\): 6 terms.
Suppose you stop the alternating harmonic series \(\sum \frac{(-1)^{n+1}}{n}\) after \(N = 9\) terms. The first omitted term has size \(b_{10} = \frac{1}{10}\), so the estimation theorem guarantees \(|S - S_9| \le\) ____
Reveal the answer
\(|S - S_9| \le b_{10} = \frac{1}{10} = \) 0.1. Set k = 9 on the slider above: the true sum sits inside the ±0.1 bracket — and the actual gap is quite a bit smaller than the bound, which is typical.
More info — why the sum gets trapped between consecutive partial sums
Watch two consecutive stops in the demo. Going from \(S_N\) to \(S_{N+1}\) adds \(\pm b_{N+1}\), jumping to the other side of the limit; going one further adds \(\mp b_{N+2}\), jumping back — but not all the way, because \(b_{N+2} \le b_{N+1}\) (that's where "decreasing" earns its keep). So every later partial sum, and hence the limit \(S\), stays inside the interval between \(S_N\) and \(S_{N+1}\). Since that interval has length exactly \(b_{N+1}\), the error bound \(|S - S_N| \le b_{N+1}\) falls out for free. One caution: this bound belongs to this test — if a series isn't alternating or fails the hypotheses, you can't use \(b_{N+1}\) as an error estimate. The OpenStax section in Dive deeper below walks the same argument with pictures.
Check your understanding
Does \(\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\) converge?
Using the alternating series error bound, how many terms of \(\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\) guarantee the partial sum is within \(0.01\) of the true sum?
The series \(\displaystyle\sum_{n=1}^{\infty} (-1)^{n+1}\frac{n}{n+1}\) alternates in sign. What does the Alternating Series Test tell you?
For the alternating harmonic series, the partial sums are \(S_4 \approx 0.5833\) and \(S_5 \approx 0.7833\). What can you conclude about the true sum \(S\)?
Recap
- Alternating Series Test: if \(b_n \ge 0\) is eventually decreasing and \(b_n \to 0\), then \(\sum (-1)^{n+1} b_n\) converges. Both hypotheses are required; check them on the term sizes \(b_n\), signs factored out.
- The partial sums oscillate and trap the sum: \(S\) lies between \(S_N\) and \(S_{N+1}\) for every \(N\).
- Remainder bound: \(|R_N| = |S - S_N| \le b_{N+1}\) — the error is at most the first omitted term, so pick \(N\) with \(b_{N+1}\) below your tolerance.
- \(\sum (-1)^{n+1}/n\) converges conditionally: convergent as signed, divergent in absolute value. The test never proves absolute convergence.
- The \(b_{N+1}\) error bound is specific to series satisfying this test — don't apply it elsewhere.
Dive deeper
- OpenStax Calculus Volume 2 — 5.5 Alternating Series Study the Alternating Series Test and the remainder estimate for truncation error.
Sources
- Alternating Series Test and the Remainder Bound