This is my proof so far,

Let $\varepsilon > 0$. By CCC, there is $N > 0$, such that, if $n,m>N$, then $|\Sigma_{i=m+1}^n a_i| < \varepsilon$. We can drop the absolute values as $a_i \geq 0$, and, $a_{m+1}\geq a_{m+2}\geq ... \geq a_n$.

now I'm stuck.


marked as duplicate by DonAntonio, grand_chat, Clement C., Rob Arthan, Mark Viola Nov 7 '16 at 21:14

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  • 3
    $\begingroup$ The number of times this was already asked on the site defies the imagination. $\endgroup$ – Did Nov 7 '16 at 21:04
  • 3
    $\begingroup$ Maybe you want to see this question I opened some days ago. $\endgroup$ – Masacroso Nov 7 '16 at 21:05
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    $\begingroup$ I think that if there are more than two weeks without a post asking this question, the universe will implode... $\endgroup$ – DonAntonio Nov 7 '16 at 21:09
  • $\begingroup$ @DonAntonio: let's hope for the continued incompetence of calculus students or their teachers! $\ddot{\smile}$. $\endgroup$ – Rob Arthan Nov 7 '16 at 21:17
  • $\begingroup$ @RobArthan Internet is an excellent excuse not to use those things called books...but if that really helps them I'm all for it. $\endgroup$ – DonAntonio Nov 7 '16 at 21:19