# Finding the Sum of a Power Series

1. Nov 27, 2006

### student45

I'm trying to find the sum of this:

$$$\sum\limits_{n = 0}^\infty {( - 1)^n nx^n }$$$

This is what I have so far:

$$$\begin{array}{l} \frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {x^n } \\ \frac{1}{{(1 - x)^2 }} = \sum\limits_{n = 0}^\infty {nx^{n - 1} } = \sum\limits_{n = 1}^\infty {nx^{n - 1} } \\ \frac{x}{{(1 - x)^2 }} = \sum\limits_{n = 1}^\infty {nx^n } \\ \end{array}$$$

So how do I get the (-1)^n part in there? Any suggestions would be really helpful. Thanks.

2. Nov 27, 2006

$$\frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^{n}x^{n}$$

Last edited: Nov 27, 2006
3. Nov 27, 2006

### student45

Oh, I see.. Where exactly does that come from?

4. Nov 27, 2006

$$\frac{1}{1+x} = \frac{1}{1-(-x)} = \sum_{n=0}^{\infty} (-x)^{n} = (-1)^{n}x^{n}$$