Solution
= Solution
Convergence of $\sum a_n$ implies $a_n\to0$. Hence $0<a_n\leq1$ for all sufficiently large $n$, and then $a_n^2\leq a_n$. The comparison test, with the finitely many initial terms treated separately, proves that
$$
\boxed{\sum_{n=1}^{\infty}a_n^2<\infty}.
$$
This is the <squares of a summable positive sequence> lemma.