Answer
The series is divergent.
Work Step by Step
$\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{(2(n+1))!}{((n+1)!)^{2}}}{\frac{(2n)!}{(n!)^{2}}}|$
$=\lim\limits_{n \to \infty}\frac{2(n+1))(2n+1)}{(n+1)^{2}}$
$=\lim\limits_{n \to \infty}\frac{2(2n+1))}{(n+1)}$
$=\lim\limits_{n \to \infty}\frac{4+2/n}{1+1/n}$
$=4 \gt 1$
The series is divergent.