If $\sum_{n=1}^{\infty} a_n$ is absolutely convergent, then $\sum_{n=1}^{\infty} (a_n)^2$ is convergent [closed]

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.

Solution 1:

If $\sum _{n=1}^\infty a_n$ converges, then you know that $\lim _{n\to \infty}a_n=0$. Thus, there is some $N\in \mathbb{Z} ^+$ such that $|a_n|<1$ for $n\geq N$. For such $a_n$, $a_n^2<|a_n|$. See if you can finish it off from here. . .

Let me know if you need an additional hint.

Solution 2:

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.