Trying to study to convergence of the series $\sum_{n=1}^{\infty} \frac{ (n!)^2 4^n }{(2n)!}$ I am trying to find out whether the following series converges:
$$ \sum_{n=1}^{\infty} \frac{ (n!)^2 4^n }{(2n)!} $$
I have tried to use the Ratio Test but it gives the inconclusive case. What other method should I employ here?
 A: Method 1. As a first remark, Stirling's approximation will do the job:
$$
a_n = \frac{n!^2 4^{n}}{(2n)!} = \frac{4^n}{\binom{2n}{n}} \operatorname*{\sim}_{n\to\infty} \sqrt{\pi n}
$$
and the series trivially diverges (the general term does not even converge to zero).

Method 2. A first simpler idea (than Stirling's): recognize a Binomial coefficient. For all $n\geq 0$,
$$
4^n = (1+1)^{2n} = \sum_{k=0}^{2n} \binom{2n}{k} > \binom{2n}{n} \tag{Middle term only}
$$
so $$a_n = \frac{4^n}{\binom{2n}{n}} > 1$$
and the series trivially diverges (the general term does not even converge to zero).

Method 3. Now, if you do not want or cannot use Stirling's nor Binomial coefficients (or are just curious and wish to apply other, reusable methods), there exist stronger versions of the ratio test.
You tried the ratio test:
$$
\frac{a_{n+1}}{a_n}
= \frac{(n+1)!^2 4^{n+1}}{(2n+2)!}\frac{(2n)!}{n!^2 4^{n}}
= \frac{4(n+1)^2}{(2n+2)(2n+1)} \xrightarrow[n\to\infty]{} 1
$$
so indeed the vanilla ratio test is inconclusive.
But you can use Raabe's test:
$$
\frac{a_n}{a_{n+1}}
= \frac{(2n+2)(2n+1)}{4(n+1)^2} = 1 - \frac{1}{2n} + o\left(\frac{1}{n}\right)
$$
when $n\to\infty$, i.e. $\lim_{n\to\infty} n\left(\left\lvert\frac{a_{n}}{a_{n+1}}\right\rvert-1\right) = -\frac{1}{2} < 1$. By Raabe's test, the series diverges.
A: Notice that for any $n \in \mathbb N$, we have that:
\begin{align*}
(2n)! 
&= \color{red}{(2n)}(2n - 1)\color{red}{(2n - 2)}(2n - 3)\color{red}{(2n - 4)}(2n - 5) \cdots \color{red}{(4)}(3)\color{red}{(2)}(1) \\
&\leq \color{red}{(2n)}(2n)\color{red}{(2n - 2)}(2n - 2)\color{red}{(2n - 4)}(2n - 4) \cdots \color{red}{(4)}(4)\color{red}{(2)}(2) \\
&= (2^n n!)^2 \\
&= (n!)^2 4^n
\end{align*}
Hence, since:
$$
a_n = \frac{(n!)^2 4^n}{(2n)!} \geq 1
$$
we know that:
$$
\lim_{n \to \infty} a_n \geq \lim_{n \to \infty} 1 = 1 > 0
$$
Thus, since $\lim a_n \neq 0$, we conclude that $\sum a_n$ diverges.
A: This is not an answer but it is too long for a comment.
If we consider the more general term $$a_n = \frac{(n!)^2 k^{n}}{(2n)!} $$ and using Stirling approximation, just as Clement C. did in his answer, we have $$a_n \operatorname*{\sim}_{n\to\infty} \left(\frac k 4 \right)^n \sqrt{\pi n}$$ and $a_n$ goes through a maximum value when $n=\frac 1{2 \log\left(\frac k 4 \right)}$ and decreases to $0$ for larger values of $n$.
Similarly, looking at the ratio test
$$\frac{a_{n+1}}{a_n}
= \frac{(n+1)!^2 k^{n+1}}{(2n+2)!}\times\frac{(2n)!}{(n!)^2 k^{n}}=\frac{k (n+1)}{4 n+2}\operatorname*{\sim}_{n\to\infty}\frac k 4$$ which is conclusive for any $k<4$.
Edit
Just for your curiosity $$\sum_{n=1}^\infty\frac{(n!)^2 k^{n}}{(2n)!}=\frac{k\sqrt{4-k} +4 \sqrt{k} \sin ^{-1}\left(\frac{\sqrt{k}}{2}\right)}{
   (4-k)^{3/2}} $$
