TITLE: Prove $\sum_{n=0}^{\infty} \frac{\Gamma(n+(1/2))}{4^n(2n+1)\Gamma(n+1)}=\frac{\pi^{3/2}}{3}$
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 0}^{\infty}{\Gamma\pars{n + 1/2} \over 4^n\pars{2n + 1}\Gamma\pars{n + 1}}} =
\Gamma\pars{1 \over 2}\sum_{n = 0}^{\infty}{\pars{n - 1/2}! \over n!\pars{-1/2}!}\,{\pars{1/4}^{n} \over 2n + 1}
\\[5mm] = &\
\root{\pi}\sum_{n = 0}^{\infty}{n - 1/2 \choose n}
\,{\pars{1/4}^{n} \over 2n + 1}
\\[5mm] = &\
\root{\pi}\sum_{n = 0}^{\infty}\bracks{{- 1/2 \choose n}\pars{-1}^{n}}
\pars{1 \over 4}^{n}\int_{0}^{1}t^{2n}\,\dd t
\\[5mm] = &\
\root{\pi}\int_{0}^{1}\sum_{n = 0}^{\infty}{- 1/2 \choose n}
\pars{-\,{t^{2} \over 4\phantom{^{2}}}}^{n}\,\dd t =
2\root{\pi}\int_{0}^{1}{\dd t \over \root{4 - t^{2}}}
\\[5mm] = &\
\bbx{\pi^{3/2} \over 3} \\ &
\end{align}