<s>[INST] Prove $\sum_{n=0}^{\infty} \frac{\Gamma(n+(1/2))}{4^n(2n+1)\Gamma(n+1)}=\frac{\pi^{3/2}}{3}$ Prove $$\sum_{n=0}^{\infty} \frac{\Gamma\left(n+\frac{1}{2}\right)}{4^n\left(2n+1\right)\Gamma\left(n+1\right)}=\frac{\pi^{\frac{3}{2}}}{3}$$
The original sum is multiplied by $\frac{\sqrt{\pi}}{2}$ and so it equals $\frac{\pi^2}{6}$ but I pulled the constant out because the actual series troubles me.  I dont know how to evaluate this.  I think maybe the Gammas and $4^n$ simplify and leave some constant divide by $2n+1$ which is the familiar arctan series.  Wolfram can't help simplify it, just compute it.  Any help please?
 [/INST] Note that $$\Gamma\left(n+\frac{1}{2}\right)=\frac{(2n)!}{4^nn!}\sqrt{\pi}, \, \, \Gamma(n+1)=n!$$ and our sum  get simplified to $$\sqrt{\pi}\sum_{n=0}^{\infty}\frac{(2n)!}{16^n(2n+1) (n!)^2}=\sum_{n=0}^{\infty}\frac{\sqrt{\pi}}{16^n(2n+1)}{2n\choose n} $$ Now recall the ordinary generating function of central binomial coefficients for $|x|<\frac{1}{4}$ , that is $$\sum_{n=0}^{\infty}{2n\choose n} x^n=\frac{1}{\sqrt{1-4x}}\cdots(1)$$ now replacing $x$by $\frac{x^2}{16}$  in $(1)$ we get $$ \sum_{n=0}^{\infty}\frac{1}{16^n}{2n\choose n}x^{2n} =\frac{2}{\sqrt{4-x^2}}\cdots(2)$$ Now  we integrating  $(2)$ from $0$ to $1$ gives $$\sum_{n=0}^{\infty}\frac{1}{16^n(2n+1)}{2n\choose n} =\int_0^1\frac{2}{\sqrt{4-x^2}}=2\int_0^1\frac{d}{dx}\sin^{-1}\left(\frac{x}{2}\right)dx=2\sin^{-1}\left(\frac{x}{2}\right)\bigg|_0^1=\frac{\pi}{3}\cdots(3)$$ now we multiply by the factor $\sqrt{\pi}$  in $(3)$ giving us the desired closed form

$$\sum_{n=0}^{\infty}\frac{\sqrt{\pi}}{16^n(2n+1)}{2n\choose n} =\frac{\pi^{\frac{3}{2}}}{3}$$

</s>