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.
 A: Note that $$4-\pi = \frac{4}{3}-\frac{4}{5}+\frac{4}{7}-\ldots$$ and by the Euler transform
$$
4-\pi = \frac{2}{3} + \frac{2}{3\cdot 5}+\frac{ 2\cdot 2}{3\cdot 5\cdot 7} +\frac{2\cdot 3 \cdot 2}{3\cdot 5\cdot 7\cdot 9}+\ldots=\sum_{n=1}^{\infty}\frac{2^{n+1}n!(n-1)!}{(2n+1)!}
$$
A: 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.
A: One may recall that, by the Wallis' integrals, we have :
$$
\frac{1}{2^{n-1}\:n}\int_0^{\pi /2} \sin^{2n+1}x \, dx =\frac{(n)!(n-1)!}{(2n+1)!}2^{n+1}, \quad n=0,1,2,..., \tag1
$$ then summing $(1)$ from $n=1$ to $\infty$, we get
$$
-2\int _0^{\pi /2}\sin x\log\left(1-\frac{\sin^{2}x}{2}\right)dx=\sum_{n=1}^{\infty}\frac{(n)!(n-1)!}{(2n+1)!}2^{n+1}. \tag2
$$ The integral is easily evaluated
$$
\begin{align}
-2\int _0^{\pi /2}\sin x\log\left(1-\frac{\sin^{2}x}{2}\right)dx&=-2\int _0^{\pi /2}\sin x\log\left(\frac{1+\cos^{2}x}{2}\right)dx\\\\
&=-2\int _0^{1}\log\left(\frac{1+u^{2}}{2}\right)du\\\\
&=4\int_0^1 \frac{u^2}{1+u^2} \, du\\\\
&= 4- \pi\tag3
\end{align}
$$ which is the desired result.
