<s>[INST] Finding the Laurent series without contour integration I would like to expand the following series around $x=0$. I know the answer already (one way is by simply plugging it into Mathematica) but I would like to understand how to get to the result.
$$\frac{1}{\sqrt{1+x} - 1} = \frac{2}{x} + \frac{1}{2} - \frac{x}{8} + \frac{x^2}{16} + ...$$
I'm not too familiar with Laurent series but from what I understand, the solution is rarely found using the contour integration of Laurent's theorem - rather it usually involves some algebraic manipulation till one reaches a geometric sum. 
However, the square root is throwing me off and most introductory examples only involve polynomials of $x$. 
 [/INST] How about using the expansion (for $|x| < 1$)
$$
\sqrt{1+x} = 1 + x/2 - x^2/8 + x^3/16 - (5 x^4)/128 + (7 x^5)/256 + O(x^6)
$$ 
Hence
$$
\frac{x}{\sqrt{1+x} - 1} = \frac{1}{1/2 - x/8 + x^2/16 - (5 x^3)/128 + (7 x^4)/256 + O(x^5)}
$$
which can be expanded
$$
\frac{x}{\sqrt{1+x} - 1} =2 + x/2 - x^2/8 + x^3/16 - (5 x^4)/128 - (7 x^5)/128 + O(x^6)
$$
Now divide by $x$ and you're done
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