Prove $\sum_{n=0}^{\infty} \frac{\Gamma(n+(1/2))}{4^n(2n+1)\Gamma(n+1)}=\frac{\pi^{3/2}}{3}$

We use the Taylor series for $\arcsin$.

Begin with $$ (1-x^2)^{-1/2} = \sum_{k=0}^\infty \binom{-1/2}{k} (-1)^k x^{2k} $$ Integrate term-by-term $$ \arcsin(x) = \sum_{k=0}^\infty\binom{-1/2}{k}\frac{(-1)^k\;x^{2k+1}}{2k+1} $$ Prove (by induction) that $$ \binom{-1/2}{k} = \frac{(-1)^{k}\;\Gamma(\frac12+k)}{\sqrt{\pi}\; k!} $$ Thus $$ \arcsin(x) = \sum_{k=0}^\infty\frac{x^{2k+1}\Gamma(\frac12+k)}{\sqrt{\pi}(2k+1)k!} $$ Plug in $x=1/2$ to get $$ \arcsin \frac12 = \frac{1}{2\sqrt{\pi}}\sum_{k=0}^\infty\frac{\Gamma(\frac12+k)}{4^k(2k+1)k!} $$ Finally, $\arcsin\frac12 = \frac{\pi}{6}$. $$ \frac{\pi}{6} = \frac{1}{2\sqrt{\pi}}\sum_{k=0}^\infty\frac{\Gamma(\frac12+k)}{4^k(2k+1)k!} \\ \frac{\pi^{3/2}}{3} = \sum_{k=0}^\infty\frac{\Gamma(\frac12+k)}{4^k(2k+1)k!} $$

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

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

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}