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