# 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 $$x$$.

Now since $$\frac{1}{\sqrt{1+x}-1}=\frac{\sqrt{1+x}+1}{x},$$ we can find Taylor series of $$\sqrt{1+x}+1$$ like $$\sum_{n=0}^\infty a_n x^n$$, then $$\frac{1}{\sqrt{1+x}-1}=\frac{1}{x}\sum_{n=0}^\infty a_n x^n=\sum_{n=0}^\infty a_n x^{n-1}$$
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