Power series of $\pi$ How do I prove that
$\displaystyle{\sum_{i=1}^{\infty}{\frac{2^n(n!)^2}{(2n)!}}=1+\frac{\pi}{2}}$
 A: Challenging question. You may notice that $\frac{n!^2}{(2n)!}=\binom{2n}{n}^{-1}$, then prove that by integration by parts and induction we have
$$ \int_{0}^{\pi/2}\sin(x)^{2n-1}\,dx = \frac{4^n}{2n\binom{2n}{n}} \tag{1}$$
It follows that
$$ S=\sum_{n\geq 1}2^n\binom{2n}{n}^{-1} = \int_{0}^{\pi/2}\sum_{n\geq 1}\frac{2n}{2^n}\sin(x)^{2n-1}\,dx = \int_{0}^{\pi/2}\frac{4\sin(x)}{(2-\sin^2 x)^2}\,dx.\tag{2}$$
Now it is not difficult to check that the last integral equals $\frac{\pi+2}{2}$: for instance, by integration by parts, since it equals:
$$ S = \int_{0}^{1}\frac{4\,dt}{(1+t^2)^2}.\tag{3}$$
A: One way to deal with this is notice the factor in the summand is proportional
to a Beta function.
$$\frac{n!^2}{(2n)!} = \frac{n}{2}\frac{\Gamma(n)\Gamma(n)}{\Gamma(2n)}
= \frac{n}{2}\int_0^1 t^{n-1}(1-t)^{n-1} dt$$
This leads to
$$\sum_{n=1}^\infty \frac{2^n n!^2}{(2n)!}
= \int_0^1 \left(\sum_{n=1}^\infty n (2t(1-t))^{n-1}\right) dt
= \int_0^1 \frac{dt}{(1-2t(1-t))^2}
= 4\int_0^1 \frac{dt}{(1+(2t-1)^2)^2}
$$
Change variable to $2t-1 = \tan\theta$, the series reduces to
$$2\int_{-\pi/4}^{\pi/4} \frac{d\theta}{1+\tan^2\theta}
= \int_{-\pi/4}^{\pi/4} (1 + \cos2\theta) d\theta
= \left[\theta + \frac12\sin(2\theta)\right]_{-\pi/4}^{\pi/4}
= \frac{\pi}{2} + 1
$$
A: We have that
$$\sum_{n=1}^{\infty}\frac{2^n(n!)^2}{(2n)!}=\sum_{n=1}^{\infty}\frac{4^n (1/2)^n}{\binom{2n}{n}}=\sum_{n=0}^{\infty}\frac{4^n (1/2)^n}{\binom{2n}{n}}-1\\=Z(1/2)-1=\left(2\cdot\frac{\pi}{4}+2\right)-1=\frac{\pi}{2}+1$$
where we used the fact that for $|t|<1$,
$$Z(t)=\sum_{n=0}^{\infty}\frac{4^n t^n}{\binom{2n}{n}}=
\frac{1}{1-t}\sqrt{\frac{t}{1-t}}\arctan \sqrt{\frac{t}{1-t}}+\frac{1}{1-t}.$$ 
A proof is given by Theorem 2.1 here. 
