If $\sum\limits_{n=1}^\infty na_n$ converges , does $\sum\limits_{n=1}^\infty na_{n+1}$ converge?

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.

Solution 1:

Yes. Put $b_n=na_n$, so the question is now (see my comment on the question):

If $\displaystyle\sum_{n=1}^\infty b_n$ converges, does $\displaystyle\sum_{n=1}^\infty\frac{b_n}{n}$ converge?

Let $s_n=\sum_{k=1}^n b_n$. We get (partial summation) $$ \sum_{k=1}^n\frac{b_k}{k} =\sum_{k=1}^n\frac{s_k-s_{k-1}}{k} =\sum_{k=1}^n\Bigl(\frac1k-\frac1{k+1}\Bigr)s_k+\frac{s_n}{n+1} =\sum_{k=1}^n\frac1{k(k+1)}s_k+\frac{s_n}{n+1} $$ which converges as $n\to\infty$, because $s_k$ is bounded, so the sum is absolutely convergent.

Solution 2:

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.