Answer
$R=\infty$ ; the interval of convergence is $(-\infty,\infty)$
Work Step by Step
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\dfrac{x^{2n+1}}{(n+1)!}}{x^{2n}/n!}|$
$=0$
Hence, $R=\infty$ ; the interval of convergence is $(-\infty,\infty)$
