Answer
Always diverges
Work Step by Step
$$|\frac{a_{n+1}}{a_{n}}|=|\dfrac{\dfrac{x^{2n+1}}{(n+1)!}}{x^{2n}/n!}|$$
Take limits on both sides, we have $$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\dfrac{x^{2n+1}}{(n+1)!}}{x^{2n}/n!}|=0$$
Hence,
R=0
and
Interval of convergence: (0)
