Does $\sum\frac{4^n(n!)^2}{(2n)!}$ converge? Does $\sum\frac{4^n(n!)^2}{(2n)!}$ converge? The ratio test is inconclusive. I know that $\sum\frac{2^n(n!)^2}{(2n)!}$ converges, if that's of any help. 

Try
$$
  a_n = \frac{4^n (n!)^2}{(2n)!}
$$
$$
  \frac{a_{n+1}}{a_{n}} = 4 \frac{(n+1)^2}{(2n+1)(2n+2)} > 1
$$
conclude that $a_n$ is monotonically increasing and also $a_n > 0$ so
$$
    \lim \limits_{n \to \infty} a_n \ne 0
$$
So
$\sum a_n $ diverge.
