Q:

Simplifying the binomial expansion

I have $$-\frac{1}{2} \left[1-\frac{z}{2}+\left(\frac{z}{2}\right)^2 -\left(\frac{z}{2}\right)^3+...\right]$$
Why does this equal $-\frac{1}{2} \sum^\infty_{n=0} \left(\frac{z}{2}\right)^n$
and not $-\frac{1}{2} \sum^\infty_{n=0} (-1)^n\left(\frac{z}{2}\right)^n$?
I think it is the latter because the signs alternate.

A:

You are correct! 
$$-\frac12 \left(1-\frac{z}{2}+\left(\frac{z}{2}\right)^2\cdots\right)=-\frac12\sum_{n=0}^{\infty}(-1)^n(z/2)^n=\frac{1}{z-2}$$
Note that $(-1)^n=1$ when $n$ is even, and $(-1)^n=-1$ when $n$ is odd.  So, the sign alternates with every other term starting with $+1$ at $n=0$.  

