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Prove that $\sum_{n=1}^{\infty}\ a_n^2$ is convergent if $\sum_{n=1}^{\infty}\ a_n$ is absolutely convergent

If $\sum\limits_{n=1}^\infty |a_n|$ converges, the $\sum\limits_{n=1}^\infty (a_n)^2$ is also always convergent?

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marked as duplicate by Mike Spivey, Asaf Karagila, Thomas, Chris Eagle, Henry T. Horton Jan 28 '13 at 18:01

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Yes. What did you try to prove it? – Did Jan 28 '13 at 16:54
It's a cute question (+1) – user 1618033 Jan 28 '13 at 16:55
@Did That was the problem. Did not find a way to prove it – Favolas Jan 28 '13 at 16:56
up vote 8 down vote accepted


  1. If $\sum |a_n|<\infty$ then $a_n\to 0$

  2. If $a_n\to 0$ then after some $n$ you have $a^2_n\leq |a_n|$ (why?)

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you were faster ... (+1) – user 1618033 Jan 28 '13 at 16:59
@Chris'ssister: my excuses – Ilya Jan 28 '13 at 17:07
@Ilya Thanks for your suggestion. I now understand – Favolas Jan 28 '13 at 18:18

If $\sum |a_n|$ converges, then $|a_n|\to 0$, and thus for sufficiently large $n$, say $n>N$, $|a_n|<1$.

For such $n>N$, $|a_n|^2<|a_n|$, so $\sum |a_n|^2<\sum |a_n|$ and thus by comparison, $\sum|a_n|^2<\infty$.

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