proof of alternating series test

The series has partial sum

$$S_{2n+2}=a_1-a_2+a_3-+\mathrm{\dots }-a_{2n}+a_{2n+1}-a_{2n+2},$$

where the $a_j$’s are all nonnegative and nonincreasing.From above, we have the following:

$S_{2n+1}$$=S_{2n}+a_{2n+1};$

$S_{2n+2}$$=S_{2n}+(a_{2n+1}-a_{2n+2});$

$S_{2n+3}$$=S_{2n+1}-(a_{2n+2}-a_{2n+3})$

$=S_{2n+2}+a_{2n+3}$

Since $a_{2n+1}\ge a_{2n+2}\ge a_{2n+3}$, we have $S_{2n+1}\ge S_{2n+3}\ge S_{2n+2}\ge S_{2n}$. Moreover,

$$S_{2n+2}=a_1-(a_2-a_3)-(a_4-a_5)-\mathrm{\cdots }-(a_{2n}-a_{2n+1})-a_{2n+2}.$$

Because the $a_j$’s are nonincreasing, we have $S_n\ge 0$ for any $n$. Also, $S_{2n+2}\le S_{2n+1}\le a_1$. Thus, $a_1\ge S_{2n+1}\ge S_{2n+3}\ge S_{2n+2}\ge S_{2n}\ge 0$. Hence, the even partial sums $S_{2n}$ and the odd partial sums $S_{2n+1}$ are bounded. Also, the even partial sums $S_{2n}$’s are monotonically nondecreasing, while the odd partial sums $S_{2n+1}$’s are monotonically nonincreasing. Thus, the even and odd series both converge.

We note that $S_{2n+1}-S_{2n}=a_{2n+1}$. Therefore, the sums converge to the same limit if and only if $a_n\to 0$ as $n\to \mathrm{\infty }$. The theorem is then established.