If $\,a_n\searrow 0\,$ and $\,\sum_{n=1}^\infty a_n<\infty,\,$ does this imply that $\,n\log n\, a_n\to 0$?

A counterexample is given by $$ a_n=\frac{1}{k^2 e^{k^2}} $$ when $e^{(k-1)^2}\leq n< e^{k^2}$. Then the number of terms equal to $1/(k^2 e^{k^2})$ is bounded above by $e^{k^2}$, so $\sum_n a_n\leq \sum_k 1/k^2<\infty$, but when $n$ is slightly less than $e^{k^2}$, $$ n\log n\,a_n\approx1, $$ so $n\log n\, a_n\not\to 0$.

Actually, there is nothing specific with $\log n$: the best we can say is that $n\cdot a_n\to 0$.

Let $\left(R_n\right)_{n\geqslant 1}$ be a sequence of non-negative numbers with goes to infinity. Then there exists a non-decreasing sequence of non-negative numbers $\left(a_n\right)_{n\geqslant 1}$such that $\sum_{n=1}^{+\infty}a_n$ is finite but $\left(nR_na_n\right)_{n\geqslant 1}$ does not converge to $0$.

Indeed, let $\left(N_k\right)_{k\geqslant 1}$ be an increasing sequence of integers such that $R_{N_{k+1}} \geqslant k^2$ for all $k\geqslant 1$. If $N_k\lt n\leqslant N_{k+1}$, let $$a_n:=\frac 1{k^2N_{k+1}}.$$ Then $$ \sum_{k=1}^{+\infty}\sum_{n=N_k+1}^{N_{k+1}}a_n=\sum_{k=1}^{+\infty}\sum_{n=N_k+1}^{N_{k+1}}\frac 1{k^2N_{k+1}}= \sum_{k=1}^{+\infty}\frac{N_{k+1}-N_k}{k^2N_{k+1}}\leqslant \sum_{k=1}^{+\infty}\frac{1}{k^2}<+\infty $$ and for all $k\geqslant 1$, $$ N_{k+1}R_{N_{k+1}}a_{N_{k+1}}=\frac 1{k^2}R_{N_{k+1}}\geqslant 1. $$

If we want a strictly decreasing sequence, add to $a_n$ a term $b_n$ which decreases to $0$ and such that $\sum_{n\geqslant 1}nR_nb_n$ is finite.