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 $x$.
Since Laurent series of a given function is unique, then no matter how we find it, using the contour integration of Laurent's theorem or other methods.
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 $$\frac{1}{\sqrt{1+x}-1}=\frac{1}{x}\sum_{n=0}^\infty a_n x^n=\sum_{n=0}^\infty a_n x^{n-1}$$