Let $\sum_{n=0}^{\infty} a_n$ be a convergent series such that $a_n \downarrow 0$. Prove that, $ \lim_{n\to \infty} na_n = 0 $

Asked Nov 07 '16 at 21:02

Active Nov 07 '16 at 21:02

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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.

asked Nov 07 '16 at 21:02

Yung Citgo

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The number of times this was already asked on the site defies the imagination. – Did Nov 07 '16 at 21:04
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Maybe you want to see this question I opened some days ago. – Masacroso Nov 07 '16 at 21:05
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I think that if there are more than two weeks without a post asking this question, the universe will implode... – DonAntonio Nov 07 '16 at 21:09
@DonAntonio: let's hope for the continued incompetence of calculus students or their teachers! $\ddot{\smile}$. – Rob Arthan Nov 07 '16 at 21:17
@RobArthan Internet is an excellent excuse not to use those things called books...but if that really helps them I'm all for it. – DonAntonio Nov 07 '16 at 21:19
I actually tried looking for this answer but I couldn't find it I apologize. – Yung Citgo Nov 08 '16 at 14:09

Let $\sum_{n=0}^{\infty} a_n$ be a convergent series such that $a_n \downarrow 0$. Prove that, $ \lim_{n\to \infty} na_n = 0 $

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