Alternating series test

Formal statement

Alternating series test

A series of the form

$${\displaystyle \sum _{n=0}^{\mathrm{\infty }}(-1{)}^n{a}_n=a_0-a_1+a_2-a_3+\cdots }$$

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, i.e., ${\displaystyle |a_{n+1}|\le |a_n|}$, and
2. ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n=0}$ .

Alternating series estimation theorem

Moreover, let L denote the sum of the series, then the partial sum $S_k=\sum _{n=0}^k(-1{)}^n{a}_n$ approximates L with error bounded by the next omitted term:

$${\displaystyle \left|S_k-L\right|\le \left|S_k-S_{k+1}\right|=a_{k+1}.}$$

Proof

Suppose we are given a series of the form $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n-1}{a}_n$, where ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n=0}$ and ${\displaystyle a_n\ge a_{n+1}}$ for all natural numbers n. (The case $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n{a}_n$ follows by taking the negative.)

Proof of the alternating series test

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

The odd partial sums decrease monotonically:

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

while the even partial sums increase monotonically:

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

both because a_n decreases monotonically with n.

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

$${\displaystyle a_1-a_2=S_2\le S_{2m}\le S_{2m+1}\le S_1=a_1.}$$

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

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

$${\displaystyle S_{2m}\le L\le S_{2m+1}}$$

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 ${\displaystyle \left|S_k-L\right|\le a_{k+1}}$ by splitting into two cases.

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

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

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

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

as desired.

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

Newer error bounds

Philip Calabrese (1962) and Richard Johnsonbaugh (1979) have found tighter bounds.

Examples

A typical example

The alternating harmonic series

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

meets both conditions for the alternating series test and converges.

Monotonicity is needed

Both conditions in the test must be met 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 $S_{2n}$ we have $S_{2n}=\frac{2}{1}+\frac{2}{2}+\frac{2}{3}+\cdots +\frac{2}{n-1}$ which is twice the partial sum of the harmonic series, which is divergent. Hence the original series is divergent.

The test is sufficient, but not necessary

Leibniz test's monotonicity is not a necessary condition, thus the test itself is only sufficient, but not necessary.

Examples of nonmonotonic series that converge are:

$${\displaystyle \sum _{n=2}^{\mathrm{\infty }}\frac{(-1{)}^n}{n+(-1{)}^n}\text{and}\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{{\cos }^{2}n}{n^2}.}$$

In fact, for every monotonic series it is possible to obtain an infinite number of nonmonotonic series that converge to the same sum by permuting its terms with permutations satisfying the condition in Agnew's theorem.