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The question: Show that if $\sum_{n=1}^\infty a_n$ converges absolutely, then $\sum_{n=1}^\infty a_n^2$ converges absolutely.

I'm stuck, not sure what path to take in solving this. Can anyone provide a hint?

Thank you.

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hint: if $|a_n|<1$, then $a_n^2<|a_n|$. –  Lior B-S Nov 13 '12 at 6:39
    
Thank you, but what if it's greater than 1? –  Alti Nov 13 '12 at 6:41
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1 Answer

up vote 2 down vote accepted

HINT: Since $\sum_{n=1}^\infty a_n$ converges absolutely, $\lim_{n\to\infty}|a_n|=0$. Therefore $|a_n|<1$ for all sufficiently large $n$; why? Now use the fact that if $|a_n|<1$, then $a_n^2<|a_n|$.

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Great, thank you! –  Alti Nov 13 '12 at 6:46
    
@Alti: You’re welcome. –  Brian M. Scott Nov 13 '12 at 6:47
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