TITLE: If $(a_n)$ is a decreasing sequence of strictly positive numbers and if $\sum{a_n}$ is convergent, show that $\lim{na_n}=0$
QUESTION [21 upvotes]: 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?

REPLY [23 votes]: 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,
$$
\sum_{k=n/2}^n a_n\ge\frac{n}{2}\frac{\epsilon}{n}=\frac{\epsilon}{2}
$$
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.