Convergence of a series implies the convergence of the squares 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?
 A: It is not true if the series is not absolutely convergent. For example
$$
1 - 1 + \frac 1{\sqrt 2} - \frac 1{\sqrt 2} + \frac 1{\sqrt 3} - \frac 1{\sqrt 3} +\ldots
$$
converges (conditionally), but the series of squares does not converge.
A: The counterexample is given by $a_n = (-1)^n\frac1{\sqrt n}$.
A: 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.
A: You can take any positive series $(b_n)_{n\in\mathbb{N}}$, wich monotonously converges to $0$, but whose series is not convergent (such as $b_n = \frac{1}{n}$). Then the series
$$ a_n = (-1)^n \sqrt{b_n} $$
is convergent, but its squares are $a_n^2 = b_n$ wich diverges by assumption.
