Let $(x_n)\downarrow 0$ and $\sum x_n\to s$. Then $(n\cdot x_n)\to 0$
Solution 1:
Hint:
If $\sum_{k=1}^\infty x_k = s,$ then by the Cauchy criterion for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that if $n > N$ we have $\sum_{k= n+1}^{2n} x_k < \epsilon.$ Since $(x_n)$ is decreasing and positive,$\sum_{k= n+1}^{2n} x_k > nx_{2n}.$
Solution 2:
Here's a correct proof: apply summation by parts to $\sum a_n b_n$, where $b_n = 1$ for all $n$.