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?
Best Answer
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).