TITLE: How to find Laurent series for this real valued function $1/(1-x)$?
QUESTION [2 upvotes]: http://www.wolframalpha.com/input/?i=1%2F(1-x)+taylor
This page says that  
$$\dfrac{1}{1-x}=\sum\limits_{n=0}^\infty x^n = 1+x+x^2+x^3 +...
$$
when $|x|<1$,
and 
$$\dfrac{1}{1-x}=\sum\limits_{n=0}^\infty -x^{-(n+1)} = -\dfrac{1}{x}-\dfrac{1}{x^2}-\dfrac{1}{x^3}-...
$$
when $|x|>1$.
It's easy to verify that both equations are true by calculating the sum of the geometric series. And by using taylor series expansion formula I can only get the first equation. My question is how to derive the second equation. On the Wolfram Alpha page it says the second equation is given by Laurent series. I googled Laurent series and I learned that the coefficient of the Laurent series is defined by a line integral of some complex function. I assume Laurent series on real valued functions is just a specific version of the general statement. However I have limited knowledge in complex analysis and am not able to derive it by myself. Thank you in advance.

REPLY [4 votes]: If $\left|x\right| > 1$, then $1/\left|x\right| < 1$. Hence,
$$
\frac{1}{1 - x} = \frac{1/x}{1/x - 1} = \frac{-1}{x}\cdot\left(\frac{1}{1 - 1/x}\right).
$$
Now use the series for $\frac{1}{1 - u}$ for $\left|u\right| < 1$ with $u$ replaced by $1/x$.
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