If $a_{n} \in \mathbb{R}$ and $\sum_{n=1}^{\infty}a_{n}$ converges, does $\sum_{n=1}^{\infty}(-1)^na_{n}$ converge?

Question

I can think of some special cases. For example, if $a_{n} \geq 0$ for all $n$, then:

$$ \sum_{n=1}^{\infty}|(-1)^na_{n}| = \sum_{n=1}^{\infty}|(-1)^n||a_{n}| = \sum_{n=1}^{\infty}1.|a_{n}| = \sum_{n=1}^{\infty}a_{n} $$

So the alternating series converges.

Similarly, if all $a_{n} < 0$, then:

$$ \sum_{n=1}^{\infty}|(-1)^na_{n}| = -\sum_{n=1}^{\infty}a_{n} $$

Which again converges.

But the problem simply asks to prove or disprove it for any $a_{n}$, so I'm stuck.

Thanks.

Answer 1

$\sum(-1)^n{1\over n}$ converges but not $\sum(-1)^n(-1)^n{1\over n}$