TITLE: If $\sum_{n=1}^{\infty} a_n$ is absolutely convergent, then $\sum_{n=1}^{\infty} (a_n)^2$ is convergent
QUESTION [7 upvotes]: Let $\sum_{n=1}^{\infty} a_n$ be a series in R. Prove that if $\sum_{n=1}^{\infty} a_n$ is absolutely convergent, then $\sum_{n=1}^{\infty} (a_n)^2$ is convergent.

REPLY [5 votes]: Another way would be thinking $\displaystyle \sum^{\infty}_{n=1} a^2_n$ as a duality pair between a sequence in $l^{\infty}$ and $l^1$, since if $\displaystyle \sum^{\infty}_{n=1} a_n$ is absolutely convergent, then termwise $|a_n|$ is uniformly bounded above, which implies $\{a_n\}\in l^{\infty}$, using Hölder's inequality for the partial sum:
$$
\sum^{N}_{n=1} |a^2_n| \leq \sup_{n\leq N} |a_n| \, \sum^{N}_{n=1} |a_n|
$$
then by the uniform boundedness of both the partial sum $\displaystyle \sum^{N}_{n=1} |a_n|$ and $\sup_{n\leq N}|a_n|$ for $\forall N\in \mathbb{N}$, letting $N\rightarrow \infty$ would give you the result.