How do you add two series together How do you add the series 
$$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{2^{n}}{(z-3)^{n+1}} + \sum_{n=0}^{\infty}\frac{(z-3)^{n}}{4^{n+1}}\right)$$
?
is this right?
$$\begin{aligned}
&\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{2^{n}}{(z-3)^{n+1}} + \frac{(z-3)^{n}}{4^{n+1}}\right)\\
 =\> & \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(4^{n+1} \cdot 2^{n}) + (z-3)^{n} (z-3)^{n+1}}{(z-3)^{n+1}  4^{n+1}} \right)\\
 =\>& \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(4^{n+1} \cdot 2^{n}) +  (z-3)^{n+1+n}}{(z-3)^{n+1} 4^{n+1}}\right)\\
 =\>& \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(4^{n+1} \cdot 2^{n}) +  (z-3)^{2n+1}}{(z-3)^{n+1} 4^{n+1}}\right)\end{aligned}$$

Well, you're doing fine so far. You can do a bit better by noting that $$4^{n+1}=\left(2^2\right)^{n+1}=2^{2(n+1)}=2^{2n+2}$$ and that $$4^{n+1}\cdot(z-3)^{n+1}=\bigl(4(z-3)\bigr)^{n+1}=(4z-12)^{n+1}.$$
You could also distribute the $\frac12$ through, if you like.
I do have to wonder what you gain from combining these two series, though. In combined form, they don't give nearly as much information as readily, in exchange for a little saved space.
