Finding powerseries $\frac{1}{z^2-z}$ Can anyone help me out with finding the powerseries of:
$f(z)=\frac{1}{z^2-z}$ about $a=-1$?

Let $x=z+1$. Then $z^2-z=z(z-1)=(x-1)(x-2)$, so
$$
f(z)=\frac{1}{(x-1)(x-2)}=\frac{1}{1-x}-\frac{1}{2-x}
=\sum_{k=0}^{\infty}  x^k-\frac{1}{2}\sum_{k=0}^{\infty}\big(\frac{1}{2}\big)x^k
=\sum_{k=0}^{\infty} \bigg(1-\frac{1}{2^{k+1}}\bigg)(z+1)^k
$$
