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