Let $a_{n}$ be decreasing and positive. Then $\sum a_{k}$ converges implies $\lim _{n \rightarrow+\infty} n a_{n}=0$.
I think since $na_n$ is positive, the only thing to do is to find an upper bound for the sequence. But I don't know how to split the sequence to form the upper bound.
Let $\epsilon >0$. By Cauchy there is $N \in \mathbb N$ such that
$$a_{p+1}+a_{p+2}+...+a_{p+n} < \epsilon/2$$
for $n,p \ge N.$
Since $(a_n)$ is decreasing, it follows that
$$n a_{p+n} <\epsilon/2$$
for $n,p \ge N.$
Hence
$$2n a_{p+n} <\epsilon$$
for $n,p \ge N.$
Therefore
$$(N+n)a_{N+n}<\epsilon$$
for $n \ge N.$
Consequence:
$$ka_k < \epsilon$$
for $k \ge 2N.$
