TITLE: If $\sum\limits_{n=1}^\infty na_n$ converges , does $\sum\limits_{n=1}^\infty na_{n+1}$ converge?
QUESTION [14 upvotes]: I ask for some help with this question:
Prove or provide counter example:
If $\sum\limits_{n=1}^\infty na_n$ converges then $\sum\limits_{n=1}^\infty na_{n+1}$ also converges.
I tries this way:
If $\sum\limits_{n=1}^\infty na_n$ converges then $na_n \to 0$, therefore $a_n \to 0$.
There are 3 possible cases:
1) If $a_n >0 $ and $a_n$ is monotonic decreasing sequence then $na_{n+1}<na_n$ and $\sum_{n=1}^\infty na_{n+1}$ converges by Comparison Test.
2) If $a_n >0 $ and $a_n$ is not monotonic decreasing sequence : it is not possible that $a_{n+1}>a_n$ because in this case $a_n \to \infty$, therefore it must be $a_{n+1} \le a_n$ and $\sum_{n=1}^\infty na_{n+1}$ converges by Comparison Test.
3) If $a_n$ is sign-alternating series. There I have a problem to find a solution.
Thanks.

REPLY [5 votes]: Using the abelian and tauberian theorem seems reasonable. Let 
$$f(z)=\sum a_nz^n$$ 
Now, if 
$$\sum n a_n$$ 
converges, then 
$$f'(z) \to \sum n a_n$$
when $z \to 1^{-}$ (abelian theorem). Then 
$$\int_0^{z}f'(u)du$$ 
tends to a definite value when $z \to 1^{-}.$ Recall that $a_n=o(1/n)$. The tauberian theorem then asserts that 
$$f(z) \to \sum a_n$$
when $z \to 1^{-}$ in such a case, and hence we are done.