Answer
Converges
Work Step by Step
Consider $a_n=\dfrac{(2^n)(n!)(n!)}{(2n)!}$
Now, $l=\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n}} |=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(2^{n+1})((n+1)!)((n+1)!)}{(2(n+1))!}}{\dfrac{(2^n)(n!)(n!)}{(2n)!}}|$
Thus, we have $l=(2) \lim\limits_{n \to \infty}|\dfrac{(n+1)^2}{4n^2+2n+4n+2}|=\lim\limits_{n \to \infty}|\dfrac{n^2+1+2n}{4n^2+6n+2}|$
or, $=(2)\lim\limits_{n \to \infty}|\dfrac{1+1/n^2+2/n}{4+6/n+2/n^2}|$
So, $l=\dfrac{1}{2} \lt 1$
Hence, the series Converges by the ratio test.