## Calculus: Early Transcendentals 8th Edition

$\Sigma a^{2}_{n}$ also converges by the comparison test.
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.