Answer
The power series: $\sum_{n=0}^\infty (-\frac{1}{2^{n+1}}-(-1)^n)x^n$
Interval of convergence: $(-1,1)$
Work Step by Step
The function of $f(x)$ can be rewritten as $f(x)=\frac{(x+1)-(x-2)}{(x-2)(x+1)}=\frac{1}{x-2}-\frac{1}{x+1}=-\frac{1}{2}\cdot \frac{1}{1-x/2}-\frac{1}{1+x}$
Find the power series for each fraction.
We need the following power series:
$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$ where $|x|<1$..................(1)
Replacing $x$ by $-x$ in Equation (1),
$\frac{1}{1+x}=\sum_{n=0}^\infty (-x)^n$ where $|-x|<1$
$\frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n$ where $|x|<1$..................(2)
Replacing $x$ by $x/2$ in Equation (1),
$\frac{1}{1-x/2}=\sum_{n=0}^\infty (x/2)^n$ where $|x/2|<1$
$\frac{1}{1-x/2}=\sum_{n=0}^\infty \frac{x^n}{2^n}$ where $|x|<2$
Multiplying by $-\frac{1}{2}$,
$-\frac{1}{2}\cdot \frac{1}{1-x/2}=\sum_{n=0}^\infty \frac{x^n}{-2^{n+1}}$ where $|x|<2$........(3)
Substracting Equation (3) with (2), we get
$f(x)=\sum_{n=0}^\infty (-\frac{1}{2^{n+1}}-(-1)^n)x^n$ where $|x|<1$ and $|x|<2$
$f(x)=\sum_{n=0}^\infty (-\frac{1}{2^{n+1}}-(-1)^n)x^n$ where $|x|<1$
