Counterexample to Leibniz criterion for alternating series

Question

I have this statement and I need to say if it is true or false:

Let $\{a_n\}$ be a real sequence.

$$\lim_{n\to +\infty} a_n = 0 \quad \implies \quad \sum_{n=1}^{\infty}(-1)^na_n \quad \text{converges}$$

I know, from the Leibniz criterion that:

If $a_n \to 0$, $a_n$ is decreasing and positive then $\sum_{n=1}^{\infty}(-1)^na_n$ converges

From this fact I believe that the statement is false but I couldn't come up with an infinitesimal sequence that isn't decreasing and for that reason is divergent. I tried some function with $sin(\frac{1}{n})$ without any luck.
Any help would be very appreciated, thank you!

Answer 1

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 $\{a_n\}$ are terms that go to zero but their series diverges, and instead of $1/n^2$ one may take $b_n$, where $\{b_n\}$ is any sequence such that their series converges.

Answer 2

HINT:

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

Answer 3

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

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