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?
$$\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$$
 A: The beautiful idea of Raymond Manzoni can actually be made rigorous. Consider a finite product $\prod_{n=1}^{L}$ and take its logarithm. After using duplication formula for the gamma function and telescoping, it simplifies to the following:
$$\sum_{n=1}^{L}\frac{1}{2^n}\ln\frac{\Gamma(2^n+\frac12)}{\Gamma(2^n)}=\left(1-2^{-L}\right)\ln\left(2\sqrt{\pi}\right)-2L\ln2+2\cdot\frac{1}{2^{L+1}}\ln\Gamma(2^{L+1}).$$
This is an exact relation, valid for any $L$. Now it suffices to use Stirling,
$$\frac{1}{N}\ln\Gamma(N)=\ln N-1+O\left(\frac{\ln N}{N}\right)\qquad \mathrm{as}\; N\rightarrow\infty$$
to get
$$\sum_{n=1}^{\infty}\frac{1}{2^n}\ln\frac{\Gamma(2^n+\frac12)}{\Gamma(2^n)}=\ln\left(2\sqrt{\pi}\right)+2\left(\ln 2-1\right)=\ln\frac{8\sqrt{\pi}}{e^2}.$$
So the answer is indeed $\displaystyle\frac{8\sqrt{\pi}}{e^2}$.
A: Firstly, $\Gamma\left(m+\frac{1}{2}\right) = \frac{(2m)!}{4^mm!}\sqrt{\pi}$ when $m\in\mathbb{Z}$, so if we let $m=2^n$, what you have can be written as
$$ \frac{\Gamma\left(m+\frac{1}{2}\right)}{\Gamma\left(m\right)} = \frac{\frac{(2m)!}{4^mm!}\sqrt{\pi}}{(m-1)!} = \frac{(2^{n+1})!\sqrt{\pi}}{4^{2^n}2^n(2^n-1)!^2}. $$
Perhaps someone else can see how to simplify this further because I sure can't.
A: 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 $$e^{\log(2)} = 2$$
A: I got $\quad \displaystyle 8\sqrt{\pi}\,e^{-2}$
Now let's search a proof...
I'll use the classical 'duplication formula' for $\Gamma$ :
$$\Gamma(z)\Gamma\left(z+\frac 12\right)=\sqrt{2\pi}\ 2^{1/2-2z}\,\Gamma(2z) $$
so that we have :
$$\frac{\Gamma(2^n)\Gamma\left(2^n+\frac 12\right)}{2^{2^{n+1}}\Gamma\left(2^{n+1}\right)}=2\sqrt{\pi}\ $$
and I thought at using 'kind of multiplicative telescoping' but it seems that considering telescoping of the logarithms as nicely done by O.L. is less confusing!
A: 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 
$$
\frac{8\sqrt\pi}{e^2}
$$
