Leibniz Criterion

analysis

Definition

Leibniz Criterion

According to the Leibniz criterion, an alternating series ${\sum }_{n\ge 0}(-1{)}^n{a}_n$ converges if $a_n$ is a monotonically decreasing zero sequence (i.e.,$a_n\ge a_{n+1}\ge 0$ and ${\lim }_{n\to \infty }{a}_n=0$).

Note that this is a sufficient but not necessary condition; the implication only holds in one direction.

Visual Representation

The partial sums $s_n$ of an alternating series satisfying the Leibniz criterion “oscillate” toward the limit, with even partial sums decreasing and odd partial sums increasing.

Proof

Leibniz Criterion

Let ${\sum }_{n\ge 0}(-1{)}^n{a}_n$ be an alternating series and $s_n$ be its partial sum. Suppose $a_n$ is monotonically decreasing and ${\lim }_{n\to \infty }{a}_n=0$.

We examine the subsequences of partial sums:

$$\begin{array}{rl}{s}_{2n+1}& =(a_0-a_1)+(a_2-a_3)+\cdots +(a_{2n}-a_{2n+1})\\ s_{2n}& =a_0-(a_1-a_2)-(a_3-a_4)-\cdots -(a_{2n-1}-a_{2n})\end{array}$$

Since $a_k-a_{k+1}\ge 0$, the sequence $(s_{2n+1})$ is non-decreasing and $(s_{2n})$ is non-increasing. Furthermore:

$$0\le s\ast 2n+1<s\ast 2n\le a\_0$$

Therefore, both subsequences are bounded and monotone, and thus convergent. Since $s_{2n}-s_{2n+1}=a_{2n+1}\to 0$, they converge to the same limit $S$.