Alternating Series Test: Conditions, Error Bound and Examples

The alternating series test handles series whose terms switch sign, like 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots. If the sizes of the terms shrink steadily to zero, the partial sums zigzag ever more tightly around a limit, and the series converges.

The test

Suppose b_n > 0 for all n. The series

converges if both:

1. b_{n+1} \le b_n for all n (or for all n past some point), and
2. \lim_{n\to\infty} b_n = 0.

The same holds for \sum (-1)^n b_n, which only flips every sign.

The error bound

If the conditions hold and S is the sum, the partial sum S_N is within one term of the truth:

Moreover, S lies between any two consecutive partial sums. This is the alternating series estimation theorem, and it makes alternating series easy to approximate.

Why it works

Because the b_n decrease, each step overshoots less than the previous one. The even partial sums S_2, S_4, S_6, \ldots increase, the odd ones S_1, S_3, S_5, \ldots decrease, and every even sum stays below every odd sum. Both sequences are bounded and monotone, so they converge, and since S_{2k+1} - S_{2k} = b_{2k+1} \to 0, they converge to the same limit.

Worked examples

Example 1: the alternating harmonic series. For \sum \frac{(-1)^{n+1}}{n}, b_n = \frac{1}{n} is positive, decreasing and tends to 0. The series converges (to \ln 2). Since \sum \frac{1}{n} diverges, the convergence is conditional.

Example 2: failing the limit condition. For \sum (-1)^n \frac{n}{n+1}, b_n = \frac{n}{n+1} \to 1 \ne 0. The alternating series test does not apply, and the nth-term test shows the series diverges.

Example 3: checking "decreasing" with a derivative. For \sum (-1)^{n+1}\frac{\ln n}{n} (starting at n = 2), let f(x) = \frac{\ln x}{x}. Then f'(x) = \frac{1 - \ln x}{x^2} < 0 for x > e, so b_n decreases for n \ge 3, and \frac{\ln n}{n} \to 0. The series converges.

Example 4: estimating a sum. The series \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n!} = 1 - \frac{1}{2} + \frac{1}{6} - \frac{1}{24} + \cdots satisfies the test. Four terms give

with error at most b_5 = \frac{1}{120} \approx 0.0083. The true sum is 1 - \frac{1}{e} \approx 0.6321, and the actual error is about 0.0071.

Absolute vs conditional convergence

The test proves convergence but says nothing about absolute convergence. Check \sum b_n separately:

The ratio test gives L = 1 for all three, so it cannot settle any of them.

Common mistakes

1. Using the test to prove divergence. Failing the decreasing condition does not mean divergence. Only b_n \not\to 0 does, and that is the nth-term test.
2. Including the sign in b_n. b_n is the positive size of the term, with (-1)^n removed.
3. Skipping the decreasing check. It is usually true, but must be verified, sometimes with a derivative.
4. Calling a series absolutely convergent because it passed this test.

For a comparison of every convergence test and how to pick one, read the full guide to convergence tests.