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?
 A: 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, $\arcsin\frac12 = \frac{\pi}{6}$.
$$
\frac{\pi}{6} = \frac{1}{2\sqrt{\pi}}\sum_{k=0}^\infty\frac{\Gamma(\frac12+k)}{4^k(2k+1)k!}
\\
\frac{\pi^{3/2}}{3} = \sum_{k=0}^\infty\frac{\Gamma(\frac12+k)}{4^k(2k+1)k!}
$$
A: 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

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

A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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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}
