I have to derive this series, as in reproduce it not differentiate.
Here's what I have:
I write $\frac{1}{4z-z^2}$ as $\frac{1}{z}\frac{1}{4-z}$ and then divide $1/4$ out of the second term, so $$ \frac{1}{4z}\frac{1}{1-z/4} $$ and expand the second term. $$ \frac{1}{4z}\sum_{n=0}^\infty \frac{z^n}{4^n} $$ then multiplying the $\frac{1}{4z}$ term through and pulling off the first term i get $$ \frac{1}{4z} + \sum_{n=1}^\infty \frac{z^{n-2}}{4^n} $$ what did i do wrong?
$$\frac{1}{4z}\sum\limits_{n=0}^\infty \frac{z^n}{4^n}=\sum\limits_{n=0}^\infty \frac{z^{n-1}}{4^{n+1}}=\frac{1}{4z}+\sum\limits_{n=1}^\infty \frac{z^{n-1}}{4^{n+1}}.$$ Reindexing the last sum by $k=n-1$ gives the correct result: $$\frac{1}{4z}+\sum\limits_{n=1}^\infty \frac{z^{n-1}}{4^{n+1}}=\frac{1}{4z}+\sum\limits_{k=0}^\infty \frac{z^{k}}{4^{k+2}}.$$