One limit. Difficult?

First i tried the following:

Let $u_{n}=\frac{(n!)^2 4^n}{(2n)!}$ . If $lim \frac {u_{n+1}}{u_{n}}=a$ and : $0<a<1$ , $lim u_{n}=0$ , $1<a$ , $u_{n}=+\infty$ . The result would turn out to be a=1 so this was inconclusive.

So I used the Stirling's Approximation:

$\huge n!=\sqrt{2\pi n} (\frac{n}{e})^n$

$\huge\lim_{n \rightarrow +\infty}\frac{(n!)^2 4^n}{(2n)!}=\lim_{n \rightarrow +\infty}\frac{(\sqrt{2\pi n}(\frac{n}{e})^n)^2 4^n}{\sqrt{4\pi n}(\frac{2n}{e})^{2n}}$

$\huge=\lim_{n \rightarrow +\infty}\frac{2\pi n (\frac {n}{e})^{2n} 2^{2n}}{2\sqrt{n}\sqrt{\pi}(\frac{2n}{e})^{2n}}=\lim_{n \rightarrow +\infty}\frac{2^{2n+1}n\pi(\frac {n}{e})^{2n}}{2\sqrt{n}\sqrt{\pi}(\frac{2n}{e})^{2n}}$

$\huge=\lim_{n \rightarrow +\infty}\frac{2^{2n}\sqrt{n}\sqrt{\pi}(\frac {n}{e})^{2n}}{(\frac{2n}{e})^{2n}}=\lim_{n \rightarrow +\infty}\frac{2^{2n}\sqrt{n}\sqrt{\pi}\frac {n^{2n}}{e^{2n}}}{\frac{2^{2n}n^{2n}}{e^{2n}}}$

$\huge=\lim_{n \rightarrow +\infty}\frac{2^{2n}e^{2n}\sqrt{n}\sqrt{\pi}\frac {n^{2n}}{e^{2n}}}{{2^{2n}n^{2n}}}=\lim_{n \rightarrow +\infty}\frac{4^{n}n^{2+\frac{1}{2}}\sqrt{\pi}}{{2^{2n}n^{2n}}}$

$\huge=\lim_{n \rightarrow +\infty}\frac{4^{n}n^{\frac{4n+2}{2}}\sqrt{\pi}}{{2^{2n}n^{2n}}}=\lim_{n \rightarrow +\infty}\frac{2^{2n}\sqrt{\pi}\sqrt{n}}{{2^{2n}}}$

$\huge=\sqrt{\pi}\lim_{n \rightarrow +\infty}\sqrt{n}=\sqrt{\pi}\times +\infty=+\infty$