Why? I was trying to prove that sum(a_k) converges absolutely then sum{(a_k)^2} converges. And the solution assumed this.

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    $\begingroup$ If $\sum a_k$ converges, then $a_k\rightarrow 0$. This is so because $a_k=S_k-S_{k-1}$ where $S_k$ is the $k^{\rm th}$ partial sum of the series ($(S_k)$ is Cauchy). $\endgroup$ – David Mitra Dec 23 '13 at 11:57
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    $\begingroup$ The title isn't supposed to be the first line of your question. How would you like it if the title of movies was the first line in the movie? $\endgroup$ – Git Gud Dec 23 '13 at 11:58
  • $\begingroup$ As for the thing you're trying to prove, if you look around the site you'll find it. Hint: I've proven it here. $\endgroup$ – Git Gud Dec 23 '13 at 12:02

If any series converges, absolutely or not, the general term sequence's limit is zero, since:

$$S=\sum_{n=1}^\infty a_n\implies a_n=S_n-S_{n-1}\xrightarrow[n\to\infty]{}S-S=0$$


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