Answer
$R=2$ ; the interval of convergence is $(-1,3]$
Work Step by Step
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(x-1)^{n+1}}{(2(n+1)-1)2^{n+1}}}{(x-1)^{n}/(2n-1)2^{n}}|$
$=\frac{x-1}{2}$
Hence, $R=2$ ; the interval of convergence is $(-1,3]$
