Finding the Sum of a Power Series

student45
I'm trying to find the sum of this:

<br /> \[<br /> \sum\limits_{n = 0}^\infty {( - 1)^n nx^n } <br /> \]<br />

This is what I have so far:

<br /> \[<br /> \begin{array}{l}<br /> \frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {x^n } \\ <br /> \frac{1}{{(1 - x)^2 }} = \sum\limits_{n = 0}^\infty {nx^{n - 1} } = \sum\limits_{n = 1}^\infty {nx^{n - 1} } \\ <br /> \frac{x}{{(1 - x)^2 }} = \sum\limits_{n = 1}^\infty {nx^n } \\ <br /> \end{array}<br /> \]<br />

So how do I get the (-1)^n part in there? Any suggestions would be really helpful. Thanks.
 
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\frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^{n}x^{n}
 
Last edited:
Oh, I see.. Where exactly does that come from?
 
\frac{1}{1+x} = \frac{1}{1-(-x)} = \sum_{n=0}^{\infty} (-x)^{n} = (-1)^{n}x^{n}
 
Ah! Of course. Okay. Thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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