TITLE: Show that $\pi =4-\sum_{n=1}^{\infty }\frac{(n)!(n-1)!}{(2n+1)!}2^{n+1}$
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: Since $\displaystyle{2n\choose n}=\frac{(2n)!}{(n!)^2}~,$ your formula is obviously connected to $\displaystyle\sum_{n=1}^\infty\frac{(2x)^{2n}}{\displaystyle{2n\choose n}~n^2}~=~2~\arcsin^2x,$ 
which can be proven by integrating the Cauchy product between the Taylor series expansions of 
$~\dfrac1{\sqrt{1-x^2}}~$ and $\arcsin x,~$ where the former is deduced from Newton's binomial series, and the 
latter is obtained by integrating the former term by term.