TITLE: Convergence of a series implies the convergence of the squares
QUESTION [8 upvotes]: Suppose $a_n\in\Bbb{R}$ and $\sum a_n$ converges. Is it necessarily true that $\sum (a_n)^2$ converges?
I think the answer is yes but I was unable to prove it. It would be much easier if $\sum a_n$ was absolutely convergent. Can you help me to prove this?

REPLY [15 votes]: The answer is in general no and 
$$a_n=\frac{(-1)^n}{\sqrt{n}}$$
is a counterexample, but if you assume that $a_n>0$ then the result becomes true. In fact the series $\displaystyle\sum_n a_n$ is convergent then the sequence $(a_n)$ is convergent to $0$ and so there's $N\in\mathbb N$ such that:
$$0<a_n<1,\quad\forall n\geq N$$
hence we have
$$0<a_n^2\leq a_n<1,\quad\forall n\geq N$$
so the series $\displaystyle\sum_n a_n^2$ is convergent by comparison.