Counterexample to Leibniz criterion for alternating series

HINT:

What happens if $a_n=(-1)^n\frac1n$?

Mark Viola gave a very natural example. Here is another one, where $a_n>0$ for all $n$: take $$ a_n=\begin{cases}1/n,&\ n\ \text{ odd}\\ \ \\ 1/n^2,&\ \text{ $n$ even } \end{cases} $$

As Mark mentioned, instead of $1/n$, one may take $a_n$ where $\{a_n\}$ is any positive divergent sequence, and instead of $1/n^2$ one may take $b_n$, where $\{b_n\}$ is any positive convergent sequence.

Consider $a_n=(-1)^{n+1}\frac1n$...

$\sum_{n=1}^\infty(-1)^na_n=\sum_{n=1}^\infty-\frac1n=-\sum_{n=1}^\infty\frac1n=-\infty$...