Question: How do I prove that

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

Answer: 
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

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