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

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 ViolaNov 7 '16 at 21:14

• The number of times this was already asked on the site defies the imagination. – Did Nov 7 '16 at 21:04
• Maybe you want to see this question I opened some days ago. – Masacroso Nov 7 '16 at 21:05
• I think that if there are more than two weeks without a post asking this question, the universe will implode... – DonAntonio Nov 7 '16 at 21:09
• @DonAntonio: let's hope for the continued incompetence of calculus students or their teachers! $\ddot{\smile}$. – Rob Arthan Nov 7 '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 7 '16 at 21:19