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- Apr 14, 2013

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Show that if $\sum_{n=1}^{\infty}a_{n}$ converges, $\lim_{n \to \infty}na_{n}=0$.

That's what I tried.Could you tell me if it is right??

$\sum_{n=1}^{\infty}a_{n}$ converges,so it satisfies the Cauchy criterion,so :

$\forall m,n$ with $m>n$ it exists a $n_{0}$ such that for $m,n \geq n_{0}$:$|a_{n}+a_{n+1}+....+a_{m-1}+a_{m}|< \epsilon$ .From the Triangle inequality,we have that $|a_{n}|+|a_{n+1}|+....+|a_{m-1}|+|a_{m}|< \epsilon$ and because of the fact that $a_{n}$ is decreasing,$|a_{n}|+|a_{n+1}|+....+|a_{m-1}|+|a_{m}| \leq |a_{n}|+|a_{n}|+....+|a_{n}|+|a_{n}|=n|a_{n}|=|na_{n}| < \epsilon$.

$$|na_{n}| < \epsilon$$ ,so $\lim_{n \to \infty}na_{n}=0$