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.

$$f(x)=\frac{x-1}{x+2}=1-\frac{3}{x+2}=1-\frac{3}{2}\frac{1}{1+\frac{x}{2}}\to\\
f(x)=1-\frac{3}{2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{x}{2}\right)^n
$$
There are several ways to represent this exression. If you'd like to have everything under one sum, you can do following:
$$
1=\frac{1}{2}2\to 1=\frac{1}{2}\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^n=\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^{n+1}\\
f(x)=\sum_{n=0}^{\infty}\left(\left(\frac{1}{2}\right)^{n+1}+\frac{3}{2}(-1)^{n+1}\left(\frac{x}{2}\right)^n\right)
$$
