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:
$$ S = \int\_{0}^{1}\frac{4\,dt}{(1+t^2)^2}.\tag{3}$$
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](https://www.emis.de/journals/INTEGERS/papers/g27/g27.pdf).