1. ## Simplifying series

Note: Sorry for all these questions, got my finals in a few days.

The question
Starting with the geometric series $\frac{1}{1 - z} = 1 + z + z^2 + ... + z^n + ...$ for |z| < 1, simplify the following series for |z| < 1, (remember you can integrate or differentiate a power series term by term within its disc of convergence):
$\sum_{n = 1}^{\infty}{nz^n}$

At first I thought I could pull the n out, and write it as $\frac{n}{1 - z}$, but it doesn't really make sense. Perhaps I'm just sleep deprived, but I'm having trouble getting this solved.

Any suggestions?

2. ## Re: Simplifying series

Originally Posted by Glitch
Note: Sorry for all these questions, got my finals in a few days.

The question
Starting with the geometric series $\frac{1}{1 - z} = 1 + z + z^2 + ... + z^n + ...$ for |z| < 1, simplify the following series for |z| < 1, (remember you can integrate or differentiate a power series term by term within its disc of convergence):
$\sum_{n = 1}^{\infty}{nz^n}$

At first I thought I could pull the n out, and write it as $\frac{n}{1 - z}$, but it doesn't really make sense. Perhaps I'm just sleep deprived, but I'm having trouble getting this solved.

Any suggestions?
Hint:
$\sum_{n = 1}^{\infty}{nz^n}=z\sum_{n = 1}^{\infty}{nz^{n-1}}$

$f(z)=\sum_{n = 1}^{\infty}{z^n}=$

Then:

$f'(z)=(\sum_{n = 1}^{\infty}{z^n})'=\sum_{n = 1}^{\infty}{nz^{n-1}}=(\frac{1}{1-z})'$

$\sum_{n = 1}^{\infty}{nz^n}=z(\frac{1}{1-z})'$

3. ## Re: Simplifying series

\displaystyle \begin{align*}\sum_{n = 1}^{\infty}nz^n &= \sum_{n = 1}^{\infty}(n + 1)z^n - \sum_{n = 1}^{\infty}z^n \\ &= \frac{d}{dz}\left(\sum_{n = 1}^{\infty}z^{n + 1}\right) - \sum_{n = 1}^{\infty}z^n\end{align*}

4. ## Re: Simplifying series

Thanks guys, Also sprach Zarathustra your post really helped.