Show that $\pi =4-\sum_{n=1}^{\infty }\frac{(n)!(n-1)!}{(2n+1)!}2^{n+1}$ Show that $$\pi =4-\sum_{n=1}^{\infty }\frac{(n)!(n-1)!}{(2n+1)!}2^{n+1}$$
I found the formula of $\pi$ by using the numerical calculation but I dont have the proving. Any help would be appreciated.

Note that $$4-\pi = \frac{4}{3}-\frac{4}{5}+\frac{4}{7}-\ldots$$ and by the Euler transform
$$
4-\pi = \frac{2}{3} + \frac{2}{3\cdot 5}+\frac{ 2\cdot 2}{3\cdot 5\cdot 7} +\frac{2\cdot 3 \cdot 2}{3\cdot 5\cdot 7\cdot 9}+\ldots=\sum_{n=1}^{\infty}\frac{2^{n+1}n!(n-1)!}{(2n+1)!}
$$
