TITLE: How to prove this sequence tends to zero
QUESTION [6 upvotes]: Suppose $a_n>0$,$\sum a_n$ converges, $\{na_n\}_{\Bbb N}$ is monotonic, prove
  $$\lim na_n\ln n=0$$

My attempt so far has shown that $\{na_n\}$ is decresing: otherwise, $na_n\ge a_1\implies a_n\ge\frac{a_1}{n}\implies \sum a_n=+\infty$ which contradicts the convergence of $\sum a_n$.
My friend has got a bit further:
$(n+1)a_{n+1}< na_n\implies \frac{a_{n+1}}{a_n}<\frac{n}{n+1}$. He then let $$b_n:=\frac{a_n}{\frac{1}{n\ln n}}$$
and therefore
$$\frac{b_{n+1}}{b_n}=\frac{a_{n+1}}{a_n}\cdot\frac{(n+1)\ln(n+1)}{n\ln n}<\frac{\ln(n+1)}{\ln n}$$
However he can't prove the monotony of $b_n$ because the RHS of the inequality is too weak. In fact it only shows $\limsup \frac{b_{n+1}}{b_n}\le 1$, which is not effective here. And I can't come up with a stronger version, either.
Can you help me? Any direct help or hint will be appreciated. Thanks!

REPLY [2 votes]: I haven't enough experience yet with series, so please let me know if the following reasoning is incorrect. I may suspect it is since I don't use all the information. In that case I'll delete the answer right away.

Assume $\lim\limits na_n\ln n\ne0,$ or doesn't even exist. Then there is at least a subsequence of $\{na_n\ln n\}$ bounded below by some positive $\varepsilon$, whence we can write, for a suitable $k$, $$\sum a_n>\sum_{n>k}\frac{\varepsilon}{n\ln n}=+\infty,$$ which contradicts the convergence of $ \sum a_n $.