How to prove that if $\sum_{n=0}^{\infty}a_n$ absolutely convergent $\Rightarrow \sum_{n=0}^{\infty}(a_n)^2$ convergent

Question

I need to prove that if $\sum_{n=0}^{\infty}a_n$ absolutely convergent $\Rightarrow \sum_{n=0}^{\infty}(a_n)^2$ convergent.

Do you have any idea haw can I prove it?

Thank you!

Answer 1

$\sum_{n = 1}^{\infty} a_n$ is absolutely convergent i.e. $\sum_{n = 1}^{\infty} |a_n|$ is convergent.

We have the result from Cauchy's criterion of convergence that $\lim_{n \rightarrow \infty} a_n = 0$

Thus $\exists$ $m \in \mathbb{N}$ s.t. $|a_n| < 1$ $\forall$ $n > m$.

Thus $(a_n)^2 < |a_n|^2 < |a_n|$

Then apply comparison test.

Let $s_k = \sum_{n = 1}^{k} |a_n|$ and $t_k = \sum_{n = 1}^{k} (a_n)^2$.

Thus $t_k < s_k$

$\{s_k\}$ is convergent as $\sum_{n = 1}^{\infty} a_n$ is convergent.

So $\{t_k\}$ is also convergent.

Thus $\sum_{n = 1}^{\infty} (a_n)^2$ is convergent.

Answer 2

You can also say that, since $\sum a_k$ is absolutely convergent, $\sum |a_k|$ is convergent. Let $c:=\sum_{k=1}^\infty |a_k|$. Then $$c^2=\left(\sum_{k=1}^\infty |a_k|\right)^2>\sum_{k=1}^\infty |a_k|^2\geq \left|\sum_{k=1}^\infty a_k^2\right|$$ (note that the last term has equality if $a_k\in\mathbb{R}$). This means that the sum $S$ of $a_k^2$ is bounded by $|S|<c^2$ for a finite $c$.