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

Question

It is given that $a_{n}$ is a positive and decreasing sequence. 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 $\lim_{n \to \infty}na_{n}=0$

Answer 1

Let $s_n$ be the partial sum.

Clearly $\lim_n(s_{2n}-s_n)=0$, and $na_{2n}\leq(s_{2n}-s_n)$, so $\lim_n{na_{2n}}=0$.

Now let $\epsilon>0$, $\exists N\forall n\geq N(na_{2n}<\epsilon/4)$. For $m\geq 2N$, we may find $n$ such that $2N\leq 2n\leq m\leq 4n$. Then $ma_m\leq ma_{2n}\leq 4na_{2n}<\epsilon$ since $n\geq N$.

Therefore, $\lim_mma_m=0$.