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?
$$\begin{aligned}
&\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{2^{n}}{(z-3)^{n+1}} + \frac{(z-3)^{n}}{4^{n+1}}\right)\\
 =\> & \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(4^{n+1} \cdot 2^{n}) + (z-3)^{n} (z-3)^{n+1}}{(z-3)^{n+1}  4^{n+1}} \right)\\
 =\>& \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(4^{n+1} \cdot 2^{n}) +  (z-3)^{n+1+n}}{(z-3)^{n+1} 4^{n+1}}\right)\\
 =\>& \frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{(4^{n+1} \cdot 2^{n}) +  (z-3)^{2n+1}}{(z-3)^{n+1} 4^{n+1}}\right)\end{aligned}$$
 A: 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.
A: $$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{2^{n}}{(z-3)^{n+1}} + \sum_{n=0}^{\infty}\frac{(z-3)^{n}}{4^{n+1}}\right)$$
$$=\frac{1}{2(z-3)}\sum_{n=0}^{\infty} \left( \frac{2}{z-3}\right)  ^n + 
\frac18  \sum_{n=0}^{\infty} \left( \frac{z-3}{4}\right)  ^n   $$
evaluate the two sums using the formula for geometric series and add the results.
A: This isn't wrong but if i were you i would write it as Laurent series. $\displaystyle\frac {1}{2}\left(\sum_{n=0}^\infty \frac{2^n}{(z-3)^{n+1}}+\sum_{n=0}^\infty \frac{(z-3)^n}{4^{n+1}}\right) = \frac {1}{2}\left(\sum_{n=-1}^{-\infty} \frac{(z-3)^n}{2^{n+1}}+\sum_{n=0}^\infty \frac{(z-3)^n}{4^{n+1}}\right)$
$\displaystyle = \frac 12 \sum_{n=-\infty}^\infty a_n (z-3)^n$
where $a_n := \begin{cases}\frac{1}{2^{n+1}}\quad\text{if n<0}\\\frac{1}{4^{n+1}}\quad\text{if }n\geq 0\end{cases}$.
If you want to evaluate the sum i suggest just leaving it as it is. 
