Context: I have derived some infinite products that I think are not well known. This is the easiest of them:
$$\prod_{n=1}^{\infty}\left(\frac{4n+1}{4n-1} \right)^{4n}\left(\frac{2n^2-2n+1}{2n^2+2n+1} \right)^n=\sqrt{2}\cosh(\pi/2)\,e^{-\frac{2G}{\pi}},\tag{1}$$ where $G$ is Catalan's constant. After some searching I can't find any reference to it.
Take a look at WolframAlpha.
Question 1:
Do you know any reference to this or something like that?
Question 2:
Could you find a solution to $(1)$?
My goal being to obtain the partial product and then use asymptotics $$P_1=\prod_{n=1}^p (4n+1)^{4n}=\frac{e^{\frac{1}{24}-\frac{C}{\pi }} \,16^{p (p+1)}\, \Gamma \left(\frac{1}{4}\right)}{\sqrt{A}\,\, \Gamma \left(p+\frac{5}{4}\right)}\,\exp\left(4 \zeta ^{(1,0)}\left(-1,p+\frac{5}{4}\right) \right)$$ $$P_2=\prod_{n=1}^p (4n-1)^{4n}=\frac{e^{\frac{1}{24}+\frac{C}{\pi }} \,16^{p (p+1)}\, \Gamma \left(p+\frac{3}{4}\right)}{\sqrt{A}\,\, \Gamma \left(\frac{3}{4}\right)}\,\exp\left(4 \zeta ^{(1,0)}\left(-1,p+\frac{3}{4}\right) \right)$$ $$\color{blue}{\frac{P_1}{P_2}=\sqrt{\pi }\, e^{-\frac{2 C}{\pi }}\frac{2^{2 p+1}}{\Gamma \left(2 p+\frac{3}{2}\right)}\,\times} $$ $$\color{blue}{\exp\left(4\left(\zeta ^{(1,0)}\left(-1,p+\frac{5}{4}\right)-\zeta ^{(1,0)}\left(-1,p+\frac{3}{4}\right)\right)\right)}$$ For large values of $p$ $$4\left(\zeta ^{(1,0)}\left(-1,p+\frac{5}{4}\right)-\zeta ^{(1,0)}\left(-1,p+\frac{3}{4}\right)\right)=$$ $$2 p \log (p)+(\log (p)+1)+\frac{3}{16 p}+O\left(\frac{1}{p^2}\right)$$ $$\color{blue}{\frac{P_1}{P_2}\sim\sqrt{\pi }\, e^{1-\frac{2 C}{\pi }}\frac{2^{2 p+1}}{\Gamma \left(2 p+\frac{3}{2}\right)}\,p^{2p+1}}$$
This was the easy part.
Now, writing $$\frac{2 n^2-2 n+1}{2 n^2+2 n+1}=\frac {(n-a_1)(n-a_2) } {(n+a_1)(n+a_2) }$$ where $$a_1=\frac {1-i}2 \qquad \text{and}\qquad a_2=\frac {1+i}2$$
$$Q_a=\prod_{n=1}^p (n+a)^n=e^{A_a}$$ $$A_a=a \left(\zeta ^{(1,0)}(0,a+1)-\zeta ^{(1,0)}(0,a+p+1)\right)+$$ $$\left(\zeta ^{(1,0)}(-1,a+p+1)-\zeta ^{(1,0)}(-1,a+1)\right)$$ and using the tedious expansion of $A_a$ for large $p$ the expansion of the second product is $$\color{blue}{\prod_{n=1}^p \left(\frac{2 n^2-2 n+1}{2 n^2+2 n+1}\right)^n = e^{-(2 p+1)} \sqrt{2 (1+\cosh (\pi ))}+O\left(\frac{1}{p^2}\right)}$$
All the above make that, at the limit, the infinite product is $$\color{red}{\prod_{n=1}^{\infty}\left(\frac{4n+1}{4n-1} \right)^{4n}\left(\frac{2n^2-2n+1}{2n^2+2n+1} \right)^n=e^{-\frac{2 C}{\pi }} \sqrt{1+\cosh (\pi )}}$$ which is your formula.
In terms of asymptotics, it would be $$\sqrt{2}\, e^{-\frac{2 C}{\pi }} \cosh \left(\frac{\pi }{2}\right) \exp\left( -\frac{3}{8 p}+\frac{18}{97 p^2}-\frac{13}{188 p^3}+O\left(\frac{1}{p^4}\right)\right)$$
For $p=9$, the relative error is less than $0.01$% and for $p=18$ it reduces to $0.001$% .