Question: Let $\sum a\_n=a$ with terms non-negative. Let $ s\_n$ the n-nth partial sum.
Prove $\sum na\_n$ converge if $\sum (a-s\_n)$ converge

Answer: 
$$
\begin{align}
\sum\_{k=1}^nka\_k
&=\sum\_{k=1}^n\sum\_{j=1}^ka\_k\\
&=\sum\_{j=1}^n\sum\_{k=j}^na\_k\\
&=\sum\_{j=1}^n(s\_n-s\_{j-1})
\end{align}
$$
Taking limits, we get by [Monotone Convergence](https://en.wikipedia.org/wiki/Monotone_convergence_theorem) that
$$
\begin{align}
\sum\_{k=1}^\infty ka\_k
&=\sum\_{j=1}^\infty(a-s\_{j-1})\\
&=a+\sum\_{j=1}^\infty(a-s\_j)\\
\end{align}
$$
Simply note

$$
\sum\_n (a-s\_n)=\sum\_n \sum\_{k=n+1}^\infty a\_k = \sum\_{k=2}^\infty \sum\_{n=1}^{k-1} a\_k =\sum\_{k=2}^\infty (k-1)a\_k.
$$

Here, we can interchange the same since all commands are nonnegative.

Using the assumption that $\sum\_n a\_n$ converges, this easily implies the claim (even with if and only if).