Alternating series test

Formulation

A series of the form

where either all a_n are positive or all a_n are negative, is called an alternating series.

The alternating series test then says: if | a n | decreases monotonically and lim n → ∞ a n = 0 then the alternating series converges.

Moreover, let L denote the sum of the series, then the partial sum

approximates L with error bounded by the next omitted term:

Proof

Suppose we are given a series of the form ∑ n = 1 ∞ ( − 1 ) n − 1 a n , where lim n → ∞ a n = 0 and a n ≥ a n + 1 for all natural numbers n. (The case ∑ n = 1 ∞ ( − 1 ) n a n follows by taking the negative.)

Proof of convergence

We will prove that both the partial sums S 2 m + 1 = ∑ n = 1 2 m + 1 ( − 1 ) n − 1 a n with odd number of terms, and S 2 m = ∑ n = 1 2 m ( − 1 ) n − 1 a n with even number of terms, converge to the same number L. Thus the usual partial sum S k = ∑ n = 1 k ( − 1 ) n − 1 a n also converges to L.

The odd partial sums decrease monotonically:

while the even partial sums increase monotonically:

both because a_n decrease monotonically with n.

Moreover, since a_n are positive, S 2 m + 1 − S 2 m = a 2 m + 1 ≥ 0 . Thus we can collect these facts to form the following suggestive inequality:

Now, note that a₁ − a₂ is a lower bound of the monotonically decreasing sequence S_2m+1, the monotone convergence theorem then implies that this sequence converges as m approaches infinity. Similarly, the sequence of even partial sum converges too.

Finally, they must converge to the same number because

Call the limit L, then the monotone convergence theorem also tells us an extra information that

for any m. This means the partial sums of an alternating series also "alternates" above and below the final limit. More precisely, when there are odd (even) number of terms, i.e. the last term is a plus (minus) term, then the partial sum is above (below) the final limit.

This understanding leads immediately to an error bound of partial sums, shown below.

Proof of partial sum error bound

We would like to show | S k − L | ≤ a k + 1 by splitting into two cases.

When k = 2m+1, i.e. odd, then

When k = 2m, i.e. even, then

as desired.

Both cases rely essentially on the last inequality derived in the previous proof.

For an alternative proof using Cauchy's convergence test, see Alternating series.

For a generalization, see Dirichlet's test.