I want to find the domain of convergence of $\sum_{n=0}^\infty(z^n+\frac{1}{2^nz^n})$.
My first thought was to use the ratio test, but that doesn't yield anything fun. So, I was wondering if we could play with Laurent series? I know $\sum_{n=0}^\infty z^n=\frac{1}{1-z}$ for $|z|<1$, but for some reason I am having trouble with $\sum_{n=0}^\infty \frac{1}{2^nz^n}$.
I know that $\frac{1}{1-z}=-\sum_{n=1}^\infty\frac{1}{z^n}$ for $|z|>1$. So, if I substitute $\frac{1}{2}z$ in for $z$ I get: $\frac{1}{1-\frac{z}{2}}=-\sum_{n=1}^\infty(\frac{1}{2z})^n$. So, after this I suppose I would get, for the series I just calculated, that $|z|>\frac{1}{2}$, and so combining with the first part, we get that $\frac{1}{2}<|z|<1$.
So, could I say that the series in question, $\sum_{n=0}^\infty(z^n+\frac{1}{2^nz^n})$, is really just the Laurent series $\frac{1}{1-z}-\frac{1}{1-\frac{z}{2}}-1$ centered at $z=0$, and so it converges in the annulus $\frac{1}{2}<|z|<1$?
It is clear from your argument that the series converges for $\frac 1 2 <|z|<1$. To show that it does not converge outside this annulus just use the fact that if one series converges and another series diverges then the sum diverges. For example, $|z| \geq 1$ implies that $\sum z^{n}$ diverges whereas $\sum \frac 1 {2^{n}z^{n}}$ converges so the given series diverges.