TITLE: Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$
QUESTION [49 upvotes]: Is there a closed form for the following infinite product?
$$\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$$

REPLY [13 votes]: Since
$$
\Gamma\left(n+\frac12\right)=\frac{(2n)!}{n!4^n}\sqrt\pi\quad\text{and}\quad\Gamma(n)=(n-1)!
$$
We have
$$
\frac{\Gamma\left(n+\frac12\right)}{\Gamma(n)}=\color{#C00000}{\binom{2n}{n}}\frac{\color{#0000FF}{n}}{\color{#00A000}{4^n}}\sqrt\pi
$$
Letting $n=2^k$, and taking logs, we get the log of the partial product to be
$$
\begin{align}
&\sum_{k=1}^n\left(\color{#C00000}{\frac1{2^k}\log\left(2^{k+1}!\right)-\frac1{2^{k-1}}\log\left(2^k!\right)}\color{#00A000}{-\log(4)}\right)+\sum_{k=1}^\infty\left(\color{#0000FF}{\frac{k}{2^k}\log(2)}+\frac1{2^{k+1}}\log(\pi)\right)\\
&=\color{#C00000}{\frac1{2^n}\log\left(2^{n+1}!\right)-\log(2)}\color{#00A000}{-n\log(4)}\color{#0000FF}{+2\log(2)}+\frac12\log(\pi)\\
&\stackrel{\text{Stirling}}\sim\frac1{2^n}\left(2^{n+1}(n+1)\log(2)-2^{n+1}+\frac12\log(2^{n+2}\pi)\right)+(1-2n)\log(2)+\frac12\log(\pi)\\
&=(2n+2)\log(2)-2+\frac1{2^{n+1}}\log(2^{n+2}\pi)+(1-2n)\log(2)+\frac12\log(\pi)\\
&=\log(8)-2+\frac12\log(\pi)+\frac1{2^{n+1}}\log(2^{n+2}\pi)\\
&\stackrel{n\to\infty}\to\log(8)-2+\frac12\log(\pi)
\end{align}
$$
Thus, the limit of the product is 
$$
\frac{8\sqrt\pi}{e^2}
$$