Question: How do I prove that

$\displaystyle{\sum\_{i=1}^{\infty}{\frac{2^n(n!)^2}{(2n)!}}=1+\frac{\pi}{2}}$

Answer: 
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:
$$ S = \int\_{0}^{1}\frac{4\,dt}{(1+t^2)^2}.\tag{3}$$
One way to deal with this is notice the factor in the summand is proportional
to a [Beta function](https://en.wikipedia.org/wiki/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

$$2\int\_{-\pi/4}^{\pi/4} \frac{d\theta}{1+\tan^2\theta}
= \int\_{-\pi/4}^{\pi/4} (1 + \cos2\theta) d\theta
= \left[\theta + \frac12\sin(2\theta)\right]\_{-\pi/4}^{\pi/4}
= \frac{\pi}{2} + 1
$$