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

This uses the hints by Raymond Manzoni and Cameron Williams. I am not sure if the answer is correct, but this should give an idea of how to proceed. I believe the answer by Raymond Manzoni is correct.
$$ \dfrac{\Gamma\left(2^n+\frac{1}{2}\right)}{\Gamma\left(2^n\right)} = \dfrac{2^n(2^{n+1})!\sqrt{\pi}}{4^{2^n}(2^n)!^2}$$
Hence,
\begin{align}
\log \left( \dfrac{\Gamma\left(2^n+\frac{1}{2}\right)}{\Gamma\left(2^n\right)}\right) & = n \log(2) + \dfrac{\log(\pi)}2 - 2^{n+1} \log(2) + \log \left(\dbinom{2^{n+1}}{2^n}\right)
\end{align}
\begin{align}
S_n = \dfrac{\log \left( \dfrac{\Gamma\left(2^n+\frac{1}{2}\right)}{\Gamma\left(2^n\right)}\right)}{2^n} & = \dfrac{n \log(2)}{2^n} + \dfrac{\log(\pi)}{2^{n+1}} - 2 \log(2) + \dfrac1{2^{n}}\log \left(\dbinom{2^{n+1}}{2^n}\right)
\end{align}
Hence,
$$\sum_{n=1}^{\infty}\dfrac{n \log(2)}{2^n} = 2 \log(2)$$
$$\sum_{n=1}^{\infty}\dfrac{\log(\pi)}{2^{n+1}} = \dfrac{\log(\pi)}2$$
Recall that
$$\log \left(\dbinom{2k}k\right) \sim 2k \log(2) - \dfrac{\log(\pi)}2 - \dfrac{\log(k)}2$$
This gives us
$$\log \left(\dbinom{2 \cdot 2^n}{2^n}\right) \sim 2^{n+1} \log(2) - \dfrac{\log(\pi)}2 - n \dfrac{\log(2)}2$$
$$\dfrac{\log \left(\dbinom{2 \cdot 2^n}{2^n}\right)}{2^n} \sim 2 \log(2) - \dfrac{\log(\pi)}{2^{n+1}} - n \dfrac{\log(2)}{2^{n+1}}$$
This gives us $S_n \sim \dfrac{n}{2^{n+1}} \log(2)$ and hence I would hope (this step needs more justification and is probably wrong in fact) $$\sum_{n=1}^{\infty} S_n  \approx \log(2)$$
Hence, the answer you are looking for is approximately $$e^{\log(2)} = 2$$
