Understanding $\mathcal Z$-transforms and pole locations
The actual computation for the inverse Z transform is

$$\frac1{z-0.5}=\frac1z\cdot\frac1{1-\frac{0.5}{z}}=z^{-1}\sum_{k=0}^{\infty}(0.5)^k z^{-k}=\sum_{n(=k+1)=1}^\infty (0.5)^{n-1}z^{-n}$$

From that you see that the series is one-sided, $h(n)=0$ for $n&lt;1$. This should remove most of the infinities that you stumbled upon.



Note that this geometric series only converges for $\left|\frac{0.5}{z}\right|&lt;1$, that is $|z|&gt;0.5$. For applications one is only interested in the values of the Z transform on or close to the unit circle, $|z|\approx 1$, the eigenvalues and singular values of the sequence shift operator. The unit circle is clearly and comfortably contained in the region of convergence. 



So for instance, if the pole were at $z=2$, then the inverse Z transform would again use the Laurent series on and close to the unit circle, i.e., for $|z|&lt;2$, to give

$$\frac1{z-2}=-\frac12\,\frac1{1-\frac z2}=-\sum_{n=0}^{\infty}2^{(-n)-1} z^{-(-n)},$$

that is, $h(n)=0$ for $n&gt;0$ and $h(n)=-2^{n-1}$ for $n\le 0$