On the Convergence Criterion for Alternating Series

关于交错级数的审敛法则 苏剑林 · 2009-10-06 · Mathematics series

Let's first consider the series

$$S=\sum_{i=1}^n (-1)^{i+1}(1/i)=1-1/2+1/3-1/4+...+(-1)^{n+1}(1/n)$$

and ask about the convergence or divergence of $n->\infty$.

First, since $\lim_{n->\infty}(-1)^{n+1}(1/n)=0$, if S diverges, it must be that $S->\infty$.

Let's suppose, for the sake of argument, that this series diverges. Then

(1)
$$S=(1-1/2)+(1/3-1/4)+...+(1/{2n}-1/{2n+1})$$
Since every term inside the parentheses is positive, S should tend to $+\infty$.
(2)
$$S=1+(-1/2+1/3)+(-1/4+1/5)+...+(-1/{2n}+1/{2n+1})+1/{2n+2}$$
Since every term inside the parentheses is negative, and $\lim_{n->\infty}(1+1/{2n+2})=1$, S should tend to $-\infty$.
But S cannot tend to both $+-\infty$ at once, so our assumption must be wrong. Hence S converges.

From this we can derive a convergence criterion for such series:

A series of the form $\sum_{i=1}^{n}=(-1)^{i}a_i$ or $\sum_{i=1}^n=(-1)^{i+1}a_i$ is called an alternating series.

If a series $\sum_{i=1}^{\infty}=(-1)^{i}a_i$ satisfies:

1. $\lim_{n->\infty} a_n=0$
2. $a_i>= a_{i+1} $

then the series converges! The proof proceeds exactly as in the example above.