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
$$
\begin{align}
\sum\_{k=1}^\infty ka\_k
&=\sum\_{j=1}^\infty(a-s\_{j-1})\\
&=a+\sum\_{j=1}^\infty(a-s\_j)\\
\end{align}
$$
This is less rigorous than the other answer but more intuitive to me (and tells us a neat thing).

$\displaystyle \sum\_1^{N} (a-s\_n)=Na-\sum\_1^N(N+1-n)a\_n$. 

As $N \to \infty$, $\displaystyle Na-\sum\_1^N(N+1-n)a\_n \to -\sum\_1^{\infty}(1-n)a\_n = \sum\_1^{\infty} na\_n-a$.

Hence $\displaystyle \sum\_1^{\infty} (a-s\_n)$ converges $\displaystyle \iff \sum\_1^{\infty} na\_n$ converges.

The neat thing is that assuming convergence, $\displaystyle \sum\_1^{\infty} (a-s\_n) = \sum\_1^{\infty} na\_n-a$