Answer
$\Sigma a^{2}_{n}$ also converges by the comparison test.
Work Step by Step
Since $\Sigma a_{n}$ converges , $\lim\limits_{n \to \infty}a_{n}=0$, so there exists N such that $|a_{n}-0|\lt 1$ for all $n\gt N \to 0\leq a_{n} \lt 1$
For all $n\gt N \to 0\leq a^{2}_{n} \lt a_{n}$.
Since $\Sigma a_{n}$ converges, we know that $\Sigma a^{2}_{n}$
also converges by the comparison test.
