<s>[INST] 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.
 [/INST] An alternate approach:
\begin{align}
\frac{x-1}{x+2} &= \frac{x + 2 - 3}{x+2} = 1 - \frac{3}{2 + x} = 1 - \frac{3}{2} \, \frac{1}{1 + \frac{x}{2}} \\
&= 1 - \frac{3}{2} \, \sum_{n=0}^{\infty} (-1)^{n} \, \frac{x^{n}}{2^{n}} \\
&= 1 - \frac{3}{2} + \sum_{n=1}^{\infty} \frac{(-1)^{n}}{2^{n}} \, x^{n} \\
&= \sum_{n=0}^{\infty} a_{n} \, x^n
\end{align}
which leads to
$$ a_{0} = - \frac{1}{2} \hspace{5mm} a_{n} = \frac{3 \, (-1)^{n+1}}{2^{n}}, n \ge 1   $$
</s>