Closed form of $\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$

Question

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

Answer 1

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 $\displaystyle\frac{8\sqrt{\pi}}{e^2}$.

Answer 2

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!

Answer 3

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 $$ \frac{8\sqrt\pi}{e^2} $$

Answer 4

Catalan (formula $(60)$ in the paper E. Catalan, Sur la constante d'Euler et la fonction de Binet, Journal de Mathématiques Pures et Appliquées $3$e série, $\mathbf{1}$ ($1875$), pp. $209$–$240$) proved that $$ \prod\limits_{n = 1}^\infty {\sqrt[{2^n }]{{\frac{{\Gamma\! \left( {2^{n - 1} z + \frac{1}{2}} \right)}}{{\Gamma (2^{n - 1} z)}}}}} = \frac{{z^z {\rm e}^{ - z} 2\sqrt \pi }}{{\Gamma (z)}} $$ for any $z>0$ (by analytic continuation, it may be extended to $|\arg z|<\pi$). In particular, if $z=2$, $$ \prod\limits_{n = 1}^\infty {\sqrt[{2^n }]{{\frac{{\Gamma\! \left( {2^n + \frac{1}{2}} \right)}}{{\Gamma (2^n )}}}}} = 8{\rm e}^{ - 2} \sqrt \pi . $$

Answer 5

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 $$e^{\log(2)} = 2$$