Answer
$a_n=\Sigma_{n=1}^{\infty} \dfrac{1}{n^2}$ and the series $\Sigma a_n^2$ converges by the p-series test.
Work Step by Step
Proof: Let us suppose that $a_n=\Sigma_{n=1}^{\infty} \dfrac{1}{n^2}$
Then $\Sigma_{n=1}^{\infty} (a_n)^2=(\Sigma_{n=1}^{\infty} \dfrac{1}{n^2}) (\Sigma_{n=1}^{\infty} \dfrac{1}{n^2})=\Sigma_{n=1}^{\infty} \dfrac{1}{n^4}$
Also, by the p-series test ($\Sigma_{n=1}^{\infty} \dfrac{1}{n^p}\implies p \gt 1;$ series converges), so the series $\Sigma a_n^2$ converges because $p=4 \gt 1$.
