If $(a_n)$ is a decreasing sequence of strictly positive numbers and if $\sum{a_n}$ is convergent, show that $\lim{na_n}=0$ If $(a_n)$ is a decreasing sequence of strictly positive numbers and if $\sum{a_n}$ is convergent, show that $\lim{na_n}=0$
Since $(a_n)$ is decreasing and bounded below, by Monotonic Convergence Theorem, $(a_n)$ converges. So, there exists $N$ such that $|a_n-L|< \epsilon$ 
Since $\sum{a_n}$ is convergent, by the Divergence test, we have $\lim_n{a_n}=0$, which means there exists $N$ such that $|a_n-0|< \epsilon$
Then I get stuck at here.
I try to figure out the statement's meaning by inserting some examples, like $a_n=\dfrac{1}{n^2}$
Can anyone guide me on this question?
 A: Hint Use Cauchy's criterion for the convergence of a series. Remember the sequence is decreasing and positive.
A: Given $\epsilon >0$, there exists an $N$ large enough such that $n,m\ge N \implies \sum a_k =|\sum a_k|<\epsilon$. Then for any $n\ge N$ we have that $$0\le na_{2n}\le a_n+a_{n-1}+ \ ... \ +a_{2n-1}=\sum a_k <\epsilon,$$ thus we have $\lim na_{2n}=0$.
Now fix $\epsilon >0$ and fix $N$ such that $n\ge N \implies na_{2n}=|na_{2n}|<\epsilon/3$. Now for $n>2N$ we can find an $m\ge n$ such that $n/3 \le m\le n/2$, hence we have $|ma_{2m}| <\epsilon/3$. Since $a_k$ is a decreasing sequence then we have that $|ma_n|\le |ma_{2m}|<\epsilon/3$, thus $|na_n|\le |3ma_n|<\epsilon$, since $n\le 3m$. Therefore $\lim na_n=0$. 
Sorry if this is a bit messy. If you want a different proof or want me to add more detail, let me know. 
A: Suppose that $\lim\limits_{n\to\infty}na_n\ne0$. That is, there is an $\epsilon\gt0$ so that for any $N$, there is an $n\ge N$ so that $na_n\ge\epsilon$. Since $a_n$ is decreasing,
$$
\sum_{k=n/2}^n a_k\ge\frac{n}{2}\frac{\epsilon}{n}=\frac{\epsilon}{2}
$$
Since $na_n$ does not converge to $0$, we can find infinitely many such blocks from $n/2$ to $n$ that sum to at least $\epsilon/2$. Thus, we've shown the contrapositive: if $\lim\limits_{n\to\infty}na_n\ne0$, then $\sum\limits_{k=0}^\infty a_n$ diverges.
