Answer
The sum converges for $x\in(-1,1)=I$
Work Step by Step
$f(x)=\frac{2x-4}{x^2-4x+3}$
$=\frac{2x-4}{(x-1)(x-3)}$
$=\frac{A}{x-1}+\frac{B}{x-3}$
$2x-4=A(x-3)+B(x-1)$
Now let $x=1$
$2(1)-4=A(1-3)+B(1-1)$
$2=2A+0$
$A=1$
Again let $x=3$
$2(3)-4=A(3-3)+B(3-1)$
$2=0+2B$
$B=1$
Thus,
$\frac{2x-4}{x^2-4x+3}=\frac{1}{x-1}+\frac{1}{x-3}$
$=\frac{-1}{1-x}+\frac{1}{-3}[\frac{1}{1-\frac{x}{3}}]$
$=-\sum_{n=0}^{\infty}~x^n-\frac{1}{3}\sum_{n=0}^{\infty}~(\frac{x}{3})^n$
$=\sum_{n=0}^{\infty}(-1-\frac{1}{3^{n-1}})x^n$
The function $f$ is represented as the sum of two geometric series, the first converges for $x\in(-1,1)$ and the second converges for $x\in(-3,3)$
Thus, the sum converges for $x\in(-1,1)=I$
