Taylor expansion of $\frac{1}{2-z-z^2}$ The problem is:
Find the Taylor expansion of $f(z):= \dfrac{1}{2-z-z^2}$ on the disc $|z| < 1$
So far I have used partial fractions to obtain $f(z) = \dfrac{1}{3}\left(\dfrac{1}{1-z} + \dfrac{1}{2+z}\right)$ which I then rewrite as $f(z) = \dfrac{1}{3}\left(\dfrac{1}{1-z} + \dfrac{1}{1-((-1)(1+z))}\right)$.
My problem is that we are told specifically to find the expansion on the disc $|z| < 1$, and whilst $\dfrac{1}{1-z}$ is valid for $|z| < 1$, $\dfrac{1}{1-((-1)(1+z))}$ is valid for $|1+z| < 1$. I am a guessing that we cannot just say $f(z) = \dfrac{1}{3}\left(\displaystyle\sum_{n=0}^\infty z^n + (-1)^n(1+z)^n\right)$ but am not sure as my notes don't have any examples of this sort of difficulty.  

Hint: $\frac{1}{1-z}=1+z+z^2+\ldots$ as infinitely descending geometric progression. And $\frac{1}{2+z}=\frac{1}{2}\cdot\frac{1}{1-(-z/2)}=\ldots$.
