Power series representation of $f(x) = \frac{x-1}{x+2}$

Question

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.

Answer 1

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

$$a_0=1-\frac32=-\frac12\;,\;\;a_n=\frac{3(-1)^{n+1}}{2^{n+1}}\;,\;\;n\ge1$$

Answer 2

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

Answer 3

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 $$ a_{0} = - \frac{1}{2} \hspace{5mm} a_{n} = \frac{3 \, (-1)^{n+1}}{2^{n}}, n \ge 1 $$