# 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.

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## closed as off-topic by Jack, Jon Mark Perry, Rise, Daniel W. Farlow, choco_addictedJun 8 at 6:14

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Is this homework? What have you tried? – Hans Parshall May 2 '11 at 14:05
I tried using Comparison test to solve it, but it didn't work out. – Lindsay Duran May 2 '11 at 14:13
Why didn't it work out? I think it should work. Using the answer below and the comparison test, you should be able to show that $\sum_{n = 1}^\infty (a_n)^2$ is absolutely convergent. Is that really so far from what you want to show? – Hans Parshall May 2 '11 at 14:18

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. . .
hello can you explain how did you get $|a_n|<1$? did you use the geometric criterion? thnk you. – Neophyte Jun 24 at 21:20
That $\lim _{n\to \infty}a_n=0$ means that, for every $\varepsilon >0$, there is some $N\in \mathbb{Z}^+$ such that, whenever $n\geq N$, it follows that $|a_n|<\varepsilon$. Taking $\varepsilon =1$, we see that in particular there is some $N\in \mathbb{Z}^+$ such that $|a_n|<1$ for $n\geq N$. – Jonathan Gleason Jun 24 at 21:22
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.