Why does $ \sum\limits_{n=1}^{\infty } 2^{2n} \times \frac{(n!)^2}{n(2n+1)(2n)!} =2 $ ?

Observe that 
\[\frac{(n!)^2}{(2n+1) (2n)!}=\frac{n! n!}{(2n+1) (2n)!}=\frac{1}{(2n+1){2n \choose n}}.\]
Now the idea is to use the Taylor series expansion for $\arcsin^2 x$. Indeed 

 $$2\arcsin^2(x) = \sum_{n=1}^\infty \frac{2^{2n} x^{2n}}{n^2 \binom{2n}n}$$

Replace $x\mapsto\sqrt x$ to get

$$2\arcsin^2\left(\sqrt x\right) = \sum_{n=1}^\infty \frac{2^{2n} x^n}{n^2 \binom{2n}n}.$$

Differentiate both side:

$$\frac{2\arcsin\left(\sqrt x\right)}{\sqrt x\sqrt{1-x}} = \sum_{n=1}^\infty \frac{2^{2n} x^{n-1}}{n \binom{2n}n}.$$

Now multiply both sides by $x$ to obtain

$$\frac{2\sqrt x\arcsin\left(\sqrt x\right)}{\sqrt{1-x}} = \sum_{n=1}^\infty \frac{2^{2n} x^n}{n \binom{2n}n}.$$

Replace $x\mapsto x^2$ to get

$$\frac{2x\arcsin(x)}{\sqrt{1-x^2}} = \sum_{n=1}^\infty \frac{2^{2n} x^{2n}}{n \binom{2n}n}$$

Thus for $0<x<1$ we have
$$\sum_{n=1}^\infty \frac{2^{2n} x^{2n+1}}{n(2n+1) \binom{2n}n} = \int_0^x \frac{2s\arcsin(s)}{\sqrt{1-s^2}} \, ds.$$

Letting $x\to1^-$ we get

$$\sum_{n=1}^\infty \frac{2^{2n} (n!)^2}{n (2n+1)!} = \sum_{n=1}^\infty \frac{2^{2n}}{n(2n+1)\binom{2n}n} = \int_0^1 \frac{2x\arcsin(x)}{\sqrt{1-x^2}} \, dx $$
\[=-2\arcsin (x) (\sqrt{1-x^2})|_0^1+2\int_0^1(\sqrt{1-x^2})\frac{1}{(\sqrt{1-x^2})}dx\]
\[=0+2=2,\]
where we have applied integration by parts to compute the last integral.