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In mathematical analysis, the alternating series test is the method used to show that an alternating series is convergent when its terms (1) decrease in absolute value, and (2) approach zero in the limit. The test was used by Gottfried Leibniz and is sometimes known as Leibniz's test, Leibniz's rule, or the Leibniz criterion. The test is only sufficient, not necessary, so some convergent alternating series may fail the first part of the test.

Formal Statement

Alternating series test

A series of the form

${\displaystyle \sum _{n=0}^{\infty }(-1)^{n}a_{n}=a_{0}-a_{1}+a_{2}-a_{3}+\cdots \!}$

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

The alternating series test guarantees that an alternating series converges if the following two conditions are met:

1. ${\displaystyle |a_{n}|}$ decreases monotonically^[1], i.e., ${\displaystyle |a_{n+1}|\leq |a_{n}|}$, and
2. ${\displaystyle \lim _{n\to \infty }a_{n}=0}$

Alternating series estimation theorem

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

${\displaystyle S_{k}=\sum _{n=0}^{k}(-1)^{n}a_{n}\!}$

approximates L with error bounded by the next omitted term:

${\displaystyle \left|S_{k}-L\right\vert \leq \left|S_{k}-S_{k+1}\right\vert =a_{k+1}.\!}$

Proof

Suppose we are given a series of the form ${\displaystyle \sum _{n=1}^{\infty }(-1)^{n-1}a_{n}\!}$, where ${\displaystyle \lim _{n\rightarrow \infty }a_{n}=0}$ and ${\displaystyle a_{n}\geq a_{n+1))$ for all natural numbers n. (The case ${\displaystyle \sum _{n=1}^{\infty }(-1)^{n}a_{n}\!}$ follows by taking the negative.)^[1]

Proof of the alternating series test

We will prove that both the partial sums ${\displaystyle S_{2m+1}=\sum _{n=1}^{2m+1}(-1)^{n-1}a_{n))$ with odd number of terms, and ${\displaystyle S_{2m}=\sum _{n=1}^{2m}(-1)^{n-1}a_{n))$ with even number of terms, converge to the same number L. Thus the usual partial sum ${\displaystyle S_{k}=\sum _{n=1}^{k}(-1)^{n-1}a_{n))$ also converges to L.

The odd partial sums decrease monotonically:

${\displaystyle S_{2(m+1)+1}=S_{2m+1}-a_{2m+2}+a_{2m+3}\leq S_{2m+1))$

while the even partial sums increase monotonically:

${\displaystyle S_{2(m+1)}=S_{2m}+a_{2m+1}-a_{2m+2}\geq S_{2m))$

both because a_n decreases monotonically with n.

Moreover, since a_n are positive, ${\displaystyle S_{2m+1}-S_{2m}=a_{2m+1}\geq 0}$. Thus we can collect these facts to form the following suggestive inequality:

${\displaystyle a_{1}-a_{2}=S_{2}\leq S_{2m}\leq S_{2m+1}\leq S_{1}=a_{1}.}$

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

${\displaystyle \lim _{m\to \infty }(S_{2m+1}-S_{2m})=\lim _{m\to \infty }a_{2m+1}=0.}$

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

${\displaystyle S_{2m}\leq L\leq S_{2m+1))$

for any m. This means the partial sums of an alternating series also "alternates" above and below the final limit. More precisely, when there is an 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 the alternating series estimation theorem

We would like to show ${\displaystyle \left|S_{k}-L\right|\leq a_{k+1}\!}$ by splitting into two cases.

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

${\displaystyle \left|S_{2m+1}-L\right|=S_{2m+1}-L\leq S_{2m+1}-S_{2m+2}=a_{(2m+1)+1))$

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

${\displaystyle \left|S_{2m}-L\right|=L-S_{2m}\leq S_{2m+1}-S_{2m}=a_{2m+1))$

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.

Examples

A typical example

The alternating harmonic series

$${\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1)){n))=1-{\frac {1}{2))+{\frac {1}{3))-{\frac {1}{4))+{\frac {1}{5))-\cdots }$$

An example to show monotonicity is needed

All of the conditions in the test, namely convergence to zero and monotonicity, should be met in order for the conclusion to be true. For example, take the series

${\displaystyle {\frac {1}((\sqrt {2))-1))-{\frac {1}((\sqrt {2))+1))+{\frac {1}((\sqrt {3))-1))-{\frac {1}((\sqrt {3))+1))+\cdots }$

The signs are alternating and the terms tend to zero. However, monotonicity is not present and we cannot apply the test. Actually the series is divergent. Indeed, for the partial sum ${\displaystyle S_{2n))$ we have ${\displaystyle S_{2n}={\frac {2}{1))+{\frac {2}{2))+{\frac {2}{3))+\cdots +{\frac {2}{n-1))}$ which is twice the partial sum of the harmonic series, which is divergent. Hence the original series is divergent.

The test is only sufficient, not necessary

Leibniz test's monotonicity is not a necessary condition, thus the test itself is only sufficient, but not necessary. (The second part of the test is well known necessary condition of convergence for all series.) Examples of nonmonotonic series that converge are ${\displaystyle \sum _{n=2}^{\infty }{\dfrac {(-1)^{n)){n+(-1)^{n))))$ and ${\displaystyle \sum _{n=1}^{\infty }(-1)^{n}{\dfrac {\cos ^{2}n}{n^{2))}.}$