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Let $a_n>0$ for all $n\in\mathbb{N}$ and the series $\sum\limits_{n=1}^\infty (n+1) a_n^2$ converges. Does it imply the series $$ \sum\limits_{n=1}^\infty a_n $$ is also converges?

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  • $\begingroup$ What do you mean by "$<\infty$"? $\endgroup$ – Jacob Oct 9 '12 at 1:27
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    $\begingroup$ @Jacob That it converges to a finite number. $\endgroup$ – Ragib Zaman Oct 9 '12 at 1:27
  • $\begingroup$ I mean less than infinity or what is the same that the series converge $\endgroup$ – JORGE BARRERA Oct 9 '12 at 1:28
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Try $$a_n = \frac{1}{(n+1)\log(n+1)}.$$

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  • $\begingroup$ lol, beat me by a couple seconds... Will delete mine... $\endgroup$ – N. S. Oct 9 '12 at 2:02
  • $\begingroup$ mmm. I really don't know what to do with that $\endgroup$ – JORGE BARRERA Oct 9 '12 at 2:10
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    $\begingroup$ @JORGEBARRERA: they are suggesting you that you prove that for such $a_n$ you have $\sum(n+1)a_n^2<\infty$ and $\sum a_n=\infty$. $\endgroup$ – Martin Argerami Oct 9 '12 at 4:04
  • $\begingroup$ @JORGE BARRERA Comparison test shows the first converges and the same test shows the second diverges, hence your hypothesis do not imply $\sum_{i=1}^\infty a_n$ converges $\endgroup$ – Robert Cardona Feb 7 '13 at 7:41

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