Telescopic series multiplied by n

Question

If I have a convergent monotonic series $\sum_{n \in \mathbb{N}}a_n = S < \infty$ then I now that the corresponding telescoping series $\sum_{n=0}^{\infty}(a_n-a_{n+1})$ converges too.

How can I demonstrate that another series

\begin{equation} \sum_{n=0}^{\infty}n(a_n-a_{n+1}) \end{equation}

converges and has the same limit as $\sum_{n=0}^{\infty} a_n$?

I tried with the method of differences but I end up with

\begin{align} \lim_{N\to \infty} \sum_{n=0}^{N}n(a_n-a_{n+1}) &= -\lim_{N\to \infty}Na_{N+1} + \lim_{N\to \infty} \sum_{n=0}^{N}a_n\\ &= S -\lim_{N\to \infty}Na_{N+1} \end{align}

$\lim_{N\to \infty}Na_{N+1}$ is now an indeterminate form $\infty \times 0$ and I have to show that it is equal to $0$.

Sorry but you really think that $$\sum_{n=0}^{\infty}(a_n-a_{n+1})=\sum_{n=0}^{\infty}n(a_n-a_{n+1})$$ in general? — Did, Nov 02 '16 at 20:30
If $\sum_{n=0}^{\infty} a_n < \infty$ yes I do, why? Wait, I made a mistake in my original post. — Francesco Carzaniga, Nov 02 '16 at 20:34
By the way, you should test the statements you are trying to prove on sequences such as $a_n=1/n$ if $n$ is a power of $2$, $a_n=0$ otherwise. — Did, Nov 02 '16 at 20:39
There probably was a condition $a_n \geqslant a_{n+1}$ somewhere, but it got lost. — Daniel Fischer, Nov 02 '16 at 20:39
Why is $\sum_{n=0}^{\infty} a_n$ different from $\sum_{n=0}^{\infty} n(a_{n} - a_{n+1})$ then? — Francesco Carzaniga, Nov 02 '16 at 20:40
@DanielFischer Yes you're right the sequence is monotonic, what does this change? — Francesco Carzaniga, Nov 02 '16 at 20:50
If you could please tell me how, then it would solve my question... — Francesco Carzaniga, Nov 02 '16 at 21:00
@FrancescoCarzaniga Yes you're right the sequence is monotonic Please edit this into the question, since that's essential information which you left out, and which changes the answer. — dxiv, Nov 02 '16 at 21:31
Yes I saw that question and the relative answer, I was actually looking for a solution that did not require using the Cauchy condensation since we have not seen it in class yet and I would rather avoid having to prove it all over. Seems however it is the only possibility in this case. — Francesco Carzaniga, Nov 02 '16 at 21:42
"monotonic series" is absurd in this context, the missing hypothesis is that the sequence $(a_n)$ is monotonous. — Did, Nov 03 '16 at 11:50
"I was actually looking for a solution that did not require using the Cauchy condensation" Lucky you, Cauchy condensation is not used in the solutions on the other page. — Did, Nov 03 '16 at 11:51

Michael Biro · Answer 1 · 2016-11-02T20:27:07.377

1

Suppose $a_n = \frac{(-1)^n}{n}$. Then,

$$\sum_{n=1}^\infty n(a_n - a_{n+1}) = \sum_{n=1}^\infty n\left(\frac{(-1)^n}{n} - \frac{(-1)^{n+1}}{n+1}\right) = \sum_{n=1}^\infty (-1)^n \left(\frac{2n+1}{n+1}\right)$$

edited Nov 02 '16 at 20:27

answered Nov 02 '16 at 20:17

Michael Biro

13,757

Telescopic series multiplied by n

1 Answers1