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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.

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  • $\begingroup$ Yes it does. If $\sum{na_n}$ is convergent, this implies that for all indices greater than $N$, $\{a_i\}$ is a decreasing sequence converging to $0$. Hence, for $m>N, a_m>a_{m+1}$. $\endgroup$
    – user67803
    Commented Dec 1, 2013 at 12:06
  • $\begingroup$ @AyushKhaitan, I doubt that that is true. $a_i$'s could slightly oscillate. $\endgroup$
    – Listing
    Commented Dec 1, 2013 at 12:09
  • $\begingroup$ My statement still stands. $\endgroup$
    – user67803
    Commented Dec 1, 2013 at 12:11
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    $\begingroup$ If you rewrite the second sum in the form $\sum_{n=2}^\infty (n-1)a_n$, you see that it is equivalent to determine whether or not $\sum a_n$ must converge. This seems a bit easier to tackle. In particular, if $a_n\ge0$ then the answer is obviouslyl yes, by the comparison principle. $\endgroup$ Commented Dec 1, 2013 at 12:16
  • $\begingroup$ @Macavity If you use such an $a_n$, $\sum_{n=1}^\infty na_n$ won't converge. $\endgroup$
    – user10444
    Commented Dec 1, 2013 at 12:26

2 Answers 2

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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.

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    $\begingroup$ I do feel a little bad about posting such a complete answer to a homework question, yet it seemed that an antidote was needed to all the bad answers out there. $\endgroup$ Commented Dec 1, 2013 at 12:40
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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.

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    $\begingroup$ That is quite clever. $\endgroup$ Commented Dec 1, 2013 at 13:02

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