I am trying to expand the function $f(x) = \left( x - 1 \right)^{-1}$ about a large value of $x$. I believe I need to use a Laurent series to accomplish this. I am not familiar with complex analysis, however, and I don't know how to begin.
I know that the Laurent series is defined as $f(z)=\sum\limits_{n=-\infty}^{\infty} a_n \left( z - z_0 \right)^n$ with $a_n = \frac{1}{2 \pi i} \oint\limits_C \frac{f(z')dz'}{(z'-z_0)^{n+1}}$. I do not, however, know how to define a contour $C$ at infinity (wouldn't $z_0$ be infinite?). I also do not know how to evaluate the integral over said contour (residue theorem?).
WolframAlpha says that the expansion about infinity is $\sum\limits_{n>0}x^{-n}$. It does not, however, provide any steps or the reasoning behind this. How is this result achieved?
Hint : if $|x|>1$,
$$\frac{1}{x-1} = \frac{1}{x} \times \frac{1}{1- \frac{1}{x}} = \frac{1}{x} \sum_{n=0}^{+\infty}\frac{1}{x^n}=\sum_{n=1}^{+\infty}\frac{1}{x^n} $$