From the function $f(z)=\frac{z+9}{z^2-2z-3}$ I determined it can be rewritten as $f(z)=\frac{3}{z-3}+\frac{-2}{z+1}$.
The Taylor series for the partial fraction decomposition is expressed as $\sum\limits_{n=0}^\infty -(\frac{z}{3})^n$ + $\sum\limits_{n=0}^\infty (-1)^{n+1}2(z)^n$
My best attempt at adding the series together is $\sum\limits_{n=0}^\infty (-1)^{n+2}2(\frac{4z}{3})^n$. I derived this from the formula $f(x) + g(x) = \sum\limits_{n=0}^\infty (a_n + b_n)x^n$.
I don't remember this technique from Calculus II and III. My rationale is combine the $(-1)$ term to get $(-1)^{n+2}$ and rationalise $\frac{z}{3}$ with $z$ to get $\frac{4z}{3}$, raise it to the $^{nth}$ power and keep $2$ as a constant. Would I please be able to get further guidance?
Best to write the series as $\sum a_n z^n$ and $\sum b_n z^n$ to avoid arithmetic errors. Then we have $$ \sum_{n\geq 0} a_n z^n + \sum_{n\geq 0} b_n z^n = \sum_{n\geq 0} (a_n+b_n) z^n $$In particular, $$ \sum_{n\geq 0} -(1/3)^n z^n + \sum_{n\geq 0} 2(-1)^{n+1} z^n = \sum_{n\geq 0} \left(-(1/3)^n+2(-1)^{n+1} \right)z^n $$