closed form of $\prod_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{(-1)^{n-1}n}$ I am looking for the closed form of this product.
$$\prod_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{(-1)^{n-1}n}$$
I have sees it somewhere before but I can't remember it closed form. I remember the   Glaisher's constant it is invloved alone with $2^{7/6}$ and maybe e (exponential function constant) also.
Does anyone knows it closed form?

Let $$a_n=\left(\frac{n}{n+1}\right)^{(-1)^{n-1} n}$$  then
$$a_{2p}= \left(\frac{2p}{2 p+1}\right)^{-2 p}\qquad \text{and}\qquad a_{2p+1}=\left(\frac{2 p+1}{2 p+2}\right)^{2 p+1}$$
Now, using a CAS,
$$\prod_{p=1}^m a_{2p}=\frac{\sqrt[12]{2} \sqrt{\pi } \exp \left(-2 \zeta ^{(1,0)}(-1,m+1)+2 \zeta
   ^{(1,0)}\left(-1,m+\frac{3}{2}\right)+\frac{1}{4}\right)}{A^3 \,\Gamma
   \left(m+\frac{3}{2}\right)}$$
$$\prod_{p=1}^m a_{2p+1}=\frac{2 \sqrt[12]{2} \Gamma (m+2) \exp \left(2 \zeta
   ^{(1,0)}\left(-1,m+\frac{3}{2}\right)-2 \zeta
   ^{(1,0)}(-1,m+2)+\frac{1}{4}\right)}{A^3}$$
$$b_m=\frac 12\prod_{p=1}^m a_{2p}\prod_{p=1}^m a_{2p+1}$$ $$b_m=\frac{2^{\frac 16}\sqrt{\pi } \Gamma (m+2) \exp \left(-2 \zeta ^{(1,0)}(-1,m+1)+4
   \zeta ^{(1,0)}\left(-1,m+\frac{3}{2}\right)-2 \zeta
   ^{(1,0)}(-1,m+2)+\frac{1}{2}\right)}{A^6 \,\Gamma \left(m+\frac{3}{2}\right)}$$
$$b_m=\frac{2^{\frac 16} \sqrt{\pi }\, \Gamma (m+2)}{A^4 \,H(m)^2\,\Gamma
   \left(m+\frac{3}{2}\right)}\exp \left(4 \zeta ^{(1,0)}\left(-1,m+\frac{3}{2}\right)-2 \zeta
   ^{(1,0)}(-1,m+2)+\frac{1}{3}\right)$$ where appears the hyperfactorial function.
Taking logarithms and using Stirling like approximations and then continuing with Taylor expansions using $b_m=e^{\log(b_m)}$
$$b_m=\frac{2^{\frac 16} \sqrt \pi}{A^6}\left(1+\frac{1}{8 m}-\frac{49}{384 m^2}+\frac{127}{1024 m^3}+O\left(\frac{1}{m^4}\right) \right)$$
$$\color{blue}{\lim_{m\to \infty } \, b_m=\frac{2^{\frac 16} \sqrt \pi}{A^6}}$$ 
