Taylor series for the function $f(z) = \frac{1}{(z-5)(z-7)}$ on a disc centered at point $z_0=3$ I started by expressing the function as sum of two fractions using partial fraction decomposition to get $\frac{-1}{2(z-5)} + \frac{1}{2(z-7)}$
However I could only then end up writing that as the sum of two power series :
$$\frac{1}{10} \sum_{n=0}^\infty \frac{z^n}{3^n} - \frac{1}{14} \sum_{n=0}^\infty \frac{z^n}{7^n}$$ and I'm unsure how to write this in the form of a Taylor series.

If we are looking to construct the series around $z_0 = 3$, we could have $\lvert z - 3\rvert < 1$ or $\lvert z - 3\rvert > 1$ which would be the series inside and outside a disc of radius one. In the first case, 
\begin{align}
\frac{1/2}{z - 3 - 4} - \frac{1/2}{z - 3 - 2}&= \frac{-1/8}{1 - (z-3)/4} + \frac{1/4}{1-(z-3)/2}\\
&= -1/8\sum_{n=0}^{\infty}\Bigl(\frac{z-3}{4}\Bigr)^n+1/4\sum_{n=0}^{\infty}\Bigl(\frac{z-3}{2}\Bigr)^n\\
&= \frac{1}{2}\sum_{n=0}^{\infty}\biggl[\frac{1}{2^{n+1}}-\frac{1}{4^{n+1}}\biggr](z-3)^n
\end{align}
This was due to the definition of a geometric series
$$
\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}
$$
for $\lvert r\rvert < 1$.
If we want a power series convergent outside the disc, $\lvert z - 3\rvert > 1\iff 1/\lvert z -3\rvert < 1$.
$$
\frac{1}{2(z - 3)}\frac{1}{1 - 4/(z-3)} - \frac{1}{2(z-3)}\frac{1}{1 - 2/(z-3)}
$$
which leads to a Laurent series. What do you get for the Laurent series convergent outside of the disc?
