# Math Help - Substitution into an infinite sum (Laurent series) question

1. ## Substitution into an infinite sum (Laurent series) question

By substituting $\frac{1}{1-z}$ for z in the expansion

$\frac{1}{(1-z)^2}=\sum_{n=0}^\infty(n+1)z^n$

derive the Laurent Series representation

$\frac{1}{z^2}=\sum_{n=2}^\infty\frac{(-1)^n(n-1)}{(z-1)^n}$

I've tried substituting and rearranging using algebra, but I can't get anything even close to that answer.. I just end up with really complicated polynomials. Is there a "trick" to this that I'm missing??

2. May be this will be useful
$\frac{1}{z^2}=\frac{1}{({z+1-1})^2}=\frac{1}{(1-z)^2(1-\frac{1}{1-z})^2}$