# Counterexample to Leibniz criterion for alternating series

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!

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.

• Indeed. And in fact, the $1/n^2$ part could be replaced with any terms that lead to a convergent series. (+1) – Mark Viola Feb 1 '18 at 22:25
• Of course. I'll edit that into the answer. – Martin Argerami Feb 1 '18 at 22:27
• Thank you, this is a neat example – edo1998 Feb 1 '18 at 22:30

HINT:

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

• This was just so simple that i could not think of it. Thank you very much, i feel a bit dumb ahah – edo1998 Feb 1 '18 at 22:26
• You're welcome. And don't feel bad. Is there a learning point/takeaway, other than the counterexample itself, you can see that might be helpful? – Mark Viola Feb 1 '18 at 22:41
• No, thank you, i wasn't only able to find a counterexample! – edo1998 Feb 1 '18 at 22:47

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

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