Answer
$R=2$ ; interval of convergence is $(-2,2)$
Work Step by Step
Let $a_{n}=\frac{n!x^{n}}{1.3.5.....(2n-1)}$, then
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{(n+1)!x^{n+1}}{1.3.5.....(2(n+1)-1)}}{\frac{n!x^{n}}{1.3.5.....(2n-1)}}|$
$=\lim\limits_{n \to \infty}|\frac{n+1}{2n+1}(x)|\lt 1$
$=|\frac{x}{2}|\lt 1$
Hence, $R=2$ ; interval of convergence is $(-2,2)$
