# Laurent series expansion for polynomial

$\frac{z}{(z)(z+1)}$ about $z=2$

I split it into a partial fraction but not sure what to do with the term with (1+z).

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