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
$$
\begin{align}
\sum\_{k=1}^\infty ka\_k
&=\sum\_{j=1}^\infty(a-s\_{j-1})\\
&=a+\sum\_{j=1}^\infty(a-s\_j)\\
\end{align}
$$
It is easy to prove the following fact:

If $(a\_n) $ is decreasing sequence and $\sum\_{j=1}^{\infty } a\_j $ converges then $\lim\_{j\to \infty } ja\_j =0.$ 

Now let $r\_n =\sum\_{k=n+1}^{\infty } a\_n $ and assume that $\sum\_{i=1}^{\infty} r\_n <\infty .$ Observe that $(r\_n )$ is an decreasing sequence, hence by the fact $\lim\_{n\to\infty} nr\_n =0.$ Take any $\varepsilon >0$ and let $n\_0 $ by such a big that $nr\_n <\varepsilon $ and $\sum\_{k=n}^{\infty} r\_k <\varepsilon $ for $n\geqslant n\_0 -1$ then we have $$\sum\_{j=n\_0}^{\infty} ja\_j = n\_0 r\_{n\_0 -1} + \sum\_{k=n\_0}^{\infty } r\_k\leqslant \varepsilon +\varepsilon =2\varepsilon .$$ Thus the series $$\sum\_{j=1}^{\infty} ja\_j $$ converges.