Prove $\sum_{n=1}^{\infty}\frac{\Gamma(n+\frac{1}{2})}{(2n+1)(2n+2)(n-1)!}=\frac{(4-π)\sqrt{\pi}}{4}$ 
Prove
  $$S=\sum_{n=1}^{\infty}\frac{\Gamma(n+\frac{1}{2})}{(2n+1)(2n+2)(n-1)!}=\frac{(4-π)\sqrt{\pi}}{4}.$$

I don't know how to evaluate this problem .
At first I used partial fraction but I got divergent series, so I used 
$$\Gamma(n)=\int_0^{+\infty}t^{n-1}e^{-t}dt.$$
This yields
$$S = \sum_{n=1}^{\infty}\frac{\int_0^{+\infty}t^{n-1}e^{-t}dt}{(2n+1)(2n+2)(n-1)!}$$
Now i can exchange integral and sum.
But I don't know how to proceed.

Since
$$\Gamma\left(n+ \frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{4^nn!}$$
We want to prove:
$$\sum_{n=1}^{\infty}\frac{\frac{(2n)!\sqrt{\pi}}{2^{2n}n!}}{(2n+1)(2n+2)(n-1)!}=\frac{(4-π)\sqrt{\pi}}{4}$$
Which is equivalent to:
$$\sum_{n=1}^{\infty} \frac{{2n \choose n}n}{(2n+1)(2n+2)2^{2n}}=\frac{4-\pi}{4}$$
Consider the central binomial coefficient series:
$$\sum_{n=1}^{\infty} {2n \choose n} x^n=\frac{1}{\sqrt{1-4x}}$$
Substituting $x^2$ instead of $x$ we obtain:
$$\sum_{n=1}^{\infty} {2n \choose n} x^{2n}=\frac{1}{\sqrt{1-4x^2}}$$
Integrating twice we get:
$$\sum_{n=1}^{\infty} \frac{{2n \choose n} x^{2n+2}}{(2n+1)(2n+2)}=\frac{\sqrt{1-4x^2}+2x\sin^{-1}(2x)-1}{4}$$
(We have to make sure that the value at zero of the left hand side and its derivative is zero, like on the right)
Now, dividing by $x^2$, taking derivative and multiplying by $\frac{1}{2}x$ we obtain:
$$\sum_{n=1}^{\infty} \frac{{2n \choose n} nx^{2n}}{(2n+1)(2n+2)}=\frac{1-x\sin^{-1}(2x)-\sqrt{1-4x^2}}{4x^2}$$
Now we just plug in $x=\frac{1}{2}$ to obtain:
$$\sum_{n=1}^{\infty} \frac{{2n \choose n} n}{(2n+1)(2n+2)2^{2n}}=\frac{1-\frac{1}{2}\sin^{-1}(2\cdot \frac{1}{2})-\sqrt{1-4(\frac{1}{2})^2}}{4(\frac{1}{2})^2}=\frac{4-\pi}{4}$$
As desired.
