Prove $\sum na_n$ converge if $\sum (a-s_n)$ converge 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
 A: 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.
A: 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, $\displaystyle \sum_1^{\infty} (a-s_n) = \sum_1^{\infty} na_n-a$
A: 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).
A: $$
\begin{align}
\sum_{k=1}^nka_k
&=\sum_{k=1}^n\sum_{j=1}^ka_k\\
&=\sum_{j=1}^n\sum_{k=j}^na_k\\
&=\sum_{j=1}^n(s_n-s_{j-1})
\end{align}
$$
Taking limits, we get by Monotone Convergence that
$$
\begin{align}
\sum_{k=1}^\infty ka_k
&=\sum_{j=1}^\infty(a-s_{j-1})\\
&=a+\sum_{j=1}^\infty(a-s_j)\\
\end{align}
$$
