Answer
$R=2$ ; interval of convergence is $[\frac{1}{2},\frac{1}{2}]$
Work Step by Step
Let $a_{n}=n!(2x-1)^{n}$, then
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{(n+1)!(2x-1)^{n+1}}{n!(2x-1)^{n}}|$
$=|(n+1)(2x-1)|$
$=\infty$
Take $(2x-1)=0$
Thus, $x=\frac{1}{2}$
Hence, $R=2$ ; interval of convergence is $[\frac{1}{2},\frac{1}{2}]$
