Does $\sum\frac{(2n)!}{4^n(n!)^2}$ converge? Does the series $\sum_{n=1}^\infty \frac{(2n)!}{4^n(n!)^2}$ converge?
The ratio test is inconclusive since the limit of $a_{n+1}/a_n$ is $1$ where $a_n=\frac{(2n)!}{4^n(n!)^2}$. Also, according to this question:
limit of $\frac{(2n)!}{4^n(n!)^2}$, the limit of $a_n$ is $0$, so this gives no information. Also it seems that the integral test and the root test are not useful to this problem. This question Does $\sum\frac{4^n(n!)^2}{(2n)!}$ converge? is kind of relevant, but I couldn't get any help from this one.

$(2n)!/(n!)^2$ is the central binomial coefficient of the $2n$th row of Pascal’s triangle, which makes it greater than the average value along that row, namely $2^{2n}/(2n+1)$.  That means the terms of your series exceed $1/(2n+1)$, whose series is divergent.
