Answer
$R=\infty$ ; interval of convergence is $(-\infty,\infty)$
Work Step by Step
Let $a_{n}=\frac{n^{2}x^{n}}{2.4.6.....(2n)}=\frac{n^{2}x^{n}}{2^{n}n!}$, then
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{(n+1)^{2}x^{n+1}}{2^{n+1}(n+1)!}}{\frac{n^{2}x^{n}}{2^{n}n!}}|$
$=0\lt 1$
Hence, $R=\infty$ ; interval of convergence is $(-\infty,\infty)$
