Let $\sum_{k=1}^\infty a_n$ be convergent show that $\sum_{k=1}^\infty n(a_n-a_{n+1})$ converges Let $\sum\limits_{k=1}^\infty a_n$ be a convergent series where $a_n\geq0$ and $(a_n)$ is a monotone decreasing sequence prove that the series $\sum\limits_{k=1}^\infty n(a_n-a_{n+1})$ also converges.
What I tried :
Let $(A_n)$ be the sequence of partial sums of the series $\sum\limits_{k=1}^\infty a_n$ and $(B_n)$ be the sequence of partial sums of the series  $\sum\limits_{k=1}^\infty n(a_n-a_{n+1})$.
Since $A_n=\sum\limits_{k=1}^n a_k$ and $B_n=\sum\limits_{k=1}^n k(a_k-a_{k+1})$ we get that:
\begin{align}
B_n&=A_n-na_{n+1}\\
&=(a_1-a_{n+1})+(a_2-a_{n+1})+...+(a_n-a_{n+1})\\
&>(a_1-a_n)+(a_2-a_n)+...+(a_n-a_n)\\
&=B_{n-1}
\end{align}
we see that $(B_n)$ is a monotone increasing sequence $...(1)$
and
$B_n=A_n-na_{n+1}<A_n$ this implies that the sequence $(B_n)$ is bounded above ...(2)
Therefore (from (1) and (2)) the sequence $(B_n)$ converges so the series $\sum\limits_{k=1}^\infty n(a_n-a_{n+1})$ also converges
Is my proof correct?
 A: Some comments:


*

*The identity $B_n = A_n - n a_{n+1}$ is not quite obvious enough to state without proof. A quick induction proof would work, as would writing out the sums using ellipses (i.e. the symbol "$\ldots$") and simplifying.

*Showing $B_n < A_n$ establishes an upper bound based on $n$, which is not allowed (e.g. $B_n \le B_n$ always!). As $A_n$ (being the partial sums of a convergent series) is convergent, you can easily establish an upper bound (especially when you consider the fact that $A_n$ is increasing).

*You can more quickly establish that $B_n$ is increasing by observing that it is the sum of positive numbers.

*You could also further note that, if $\lim B_n < \lim A_n$, then $a_n$ is approximately a multiple of the harmonic series, which is divergent, thus the two series share the same sum. (This is not a criticism, just something worth noting).
A: Note that we can use $n=\sum_{k=1}^n (1)$ to write 
$$\begin{align}
\sum_{n=1}^N n(a_{n+1}-a_n)&=\sum_{n=1}^N \sum_{k=1}^n(a_{n+1}-a_n)\\\\
&=\sum_{k=1}^N \sum_{n=k}^N (a_{n+1}-a_n)\\\\
&=\sum_{k=1}^N (a_{N+1}-a_k)\\\\
&=Na_{N+1}-\sum_{k=1}^N a_k\tag1
\end{align}$$
Inasmuch as $a_n\ge 0$ monotonically decreases to $0$, and $\sum_{k=1}^n a_n<\infty$, we have $\lim_{n\to\infty }na_n=0$.  Hence, using $(1)$, we see that 
$$\begin{align}
\lim_{N\to \infty }\sum_{n=1}^N n(a_{n+1}-a_n)&=\lim_{N\to\infty}\left(Na_{N+1}-\sum_{k=1}^N a_k\right)\\\\
&=-\sum_{n=1}^\infty a_n
\end{align}$$
from which we conclude that $\sum_{n=1}^\infty n(a_{n+1}-a_n)$ converges.
