Limit with terms of convergent series problem Prove that if a positive series $\sum_{\nu =1}^\infty a_\nu$ is convergent and the sequence $(\nu a_\nu)_{\nu =1}^\infty$ is decreasing, then $\lim_{\nu\to\infty}(\nu\log\nu)a_\nu=0$.
I've been trying to prove this for days, but so far I've only managed to prove that if the limit exists, it is equal to 0.
Could someone give me a hint?

Suppose $na_n\log n$ does not tend to $0$.
 There exists $n_k$ increasing to $\infty $ such that $a_{{n_k}} n_k \log n_k \geq a$ for all $k$ for some $a>0$. Further, replacing $\{n_k\}$ be  a subequence if necessary we may assume that $$ \log (n_{k+1}-1) -\log n_k >\frac 1 2 \log n_{k+1}$$ for all $k$. Now $$\sum_n a_n=\sum_k \sum_{n_k \leq n <n_{k+1}} \frac 1 n (na_n)\geq \sum_k  \sum_{n=n_{k}}^{n_{k+1}-1} \frac 1 n (n_{k+1}a_{n_{k+1}})>\sum_k ({\log (n_{k+1}-1) -\log n_k}) (n_{k+1}a_{n_{k+1}}) $$. The general term of this series exceeds $\frac a 2$. This makes $\sum a_n$ divergent.
