What is the prove for this formula? $$\prod_{n=0}^{\infty}\frac{(3n+2)^5(3n+4)^{10}(3n+6)}{(3n+1)(3n+3)^{10}(3n+5)^5}=\frac{6^5}{\Gamma^9\left(\frac{1}{3}\right)}$$
$$\log\prod_{n=0}^{\infty}\frac{(3n+2)^5(3n+4)^{10}(3n+6)}{(3n+1)(3n+3)^{10}(3n+5)^5}=\log Q$$
$$\lim_{M \to \infty}\sum_{n=0}^{M}-\log(3n+1)+5\log(3n+2)-10\log(3n+3)+10\log(3n+4)-5\log(3n+5)+\log(3n+6)=\log Q$$
I am trying to understand this page but can not seem to figure it how do more on the above infinite poduct
$$P_p=\prod_{n=0}^{p}\frac{(3n+2)^5(3n+4)^{10}(3n+6)}{(3n+1)(3n+3)^{10}(3n+5)^5}=\frac{32 (p+2) (3 p+4)\, \Gamma \left(p+\frac{7}{3}\right)^9}{(3 p+5)^5 \,\Gamma \left(\frac{4}{3}\right)^9 \,\Gamma (p+2)^9}$$ Now, using Stirling approximation and continuing with Taylor series $$9\log \left(\frac{\Gamma \left(p+\frac{7}{3}\right)}{\Gamma (p+2)}\right)=3 \log \left({p}\right)+\frac{5}{p}-\frac{73}{18 p^2}+\frac{115}{27 p^3}-\frac{527}{108 p^4}+O\left(\frac{1}{p^5}\right)$$ $$\log \left(\frac{32 (p+2) (3 p+4)}{(3 p+5)^5}\right)=-3 \log \left({p}\right)-\log \left(\frac{81}{32}\right)-\frac{5}{p}+\frac{73}{18 p^2}-\frac{115}{27 p^3}+\frac{1573}{324 p^4}+O\left(\frac{1}{p^5}\right)$$ Adding these two, what we have is $$-\log \left(\frac{81}{32}\right)-\frac{2}{81p^4}+O\left(\frac{1}{p^5}\right)$$ Taking the exponential $$\frac{32}{81}-\frac{64}{6561 p^4}+O\left(\frac{1}{p^5}\right)$$ So, $$P_p\sim \frac {\frac{32}{81}-\frac{64}{6561 p^4}+O\left(\frac{1}{p^5}\right)}{\Gamma \left(\frac{4}{3}\right)^9 } =\frac {\frac{32}{81}-\frac{64}{6561 p^4}+O\left(\frac{1}{p^5}\right)}{\left(\frac{1}{3}\right)^9\,\Gamma \left(\frac{1}{3}\right)^9 }=\frac {7776-\frac {192}{p^4}+O\left(\frac{1}{p^5}\right)} {\Gamma \left(\frac{1}{3}\right)^9 } $$ and $7776=6 ^5$.