Power series representation of $f(x) = \frac{x-1}{x+2}$ What I did so far:
\begin{align*}
\frac{x-1}{x+2}&=(\frac{x-1}{2})\frac{1}{1-(-\frac{x}{2})} \\
&=(\frac{x-1}{2})\sum_{n=1}^\infty (-\frac{x}{2})^n\\
&= (x-1)\sum_{n=1}^\infty \frac{-x^n}{2^{n+1}}
\end{align*}
But i'm stuck there.

But then you're already done...That can't be put in any form that will be nicer imo.
You could also try
$$\frac{x-1}{x+2}=1-\frac3{x+2}=1-\frac32\frac1{1+\frac x2}=1-\frac32\sum_{n=0}^\infty\frac{(-1)^nx^n}{2^n}=\sum_{n=0}^\infty a_nx^n$$
with
$$a_0=1-\frac32=-\frac12\;,\;\;a_n=\frac{3(-1)^{n+1}}{2^{n+1}}\;,\;\;n\ge1$$
