To find the infinite sums, in general, the following approach seems to work: $\sum_{0}^{\infty}z^{n+1} = z/(1-z)$, where $|z|<1$. This is just different version from the standard equation for power series: $\sum_{0}^{\infty}a_nz^{n} = a_1/(1-z)$, with $|z|<1$.
When I do this for the following equation in the place of $z$: $z/(z+2)$, I get, as expected $z/2$. This is also what wolfram alpha gives me.
If I do this with $z/(z-2)$, I would expect to get $-x/2$ according to the same reasoning: $\sum_{0}^{\infty}(z/(z-2))^{n+1} = \frac{z/(z-2)}{1-z/(z-2)} = -z/2$, for $|z/(z-2)|<1$.
Wolfram Alpha however, says that this is not correct. I do not understand what is going wrong. What is wrong about this approach?
For $|r|<1$, $$ \sum_{n=0}^\infty r^n=\frac{1}{1-r}\tag{1} $$
Let $r=z/(z-2)$. For $|z/(z-2)|<1$, you have $$ \frac{1}{1-z/(z-2)}=\frac{z-2}{(z-2)-z}=\frac{z-2}{z-2-z}=-\frac12(z-2). $$
For your edited version, you simply have $$ \sum_{n=0}^\infty r^{n+1}=r\sum_{n=0}^\infty r^{n}=\frac{z}{z-2}\cdot(-\frac12)({z-2})=\frac{-z}{2}. $$