The Golden mean known as $\frac{1+\sqrt{5}}{2}$.
How could one show the Golden mean can be expressed as
$$ \frac{2\cdot 3\cdot 7\cdot 8\cdot 12\cdot 13\cdots}{1\cdot 4\cdot 6\cdot 9\cdot 11\cdot 14\cdots} $$
This is the limiting case of the Rogers-Ramanujan continued fraction. But how would you prove this by using properties of Gamma function?
Any help would be appreciated!
Consider the combination \begin{align}f_N=\frac{\Gamma\left(N+\frac{2}{5}\right)\Gamma\left(N+\frac{3}{5}\right)}{\Gamma\left(N+\frac{1}{5}\right)\Gamma\left(N+\frac{4}{5}\right)}= \frac{\left(5N-2\right)\cdot\left(5N-3\right)}{\left(5N-1\right)\cdot\left(5N-4\right)}f_{N-1}=\ldots=\\= \frac{\left(5N-2\right)\cdot\left(5N-3\right)}{\left(5N-1\right)\cdot\left(5N-4\right)}\times\ldots\times \frac{8\cdot 7}{9\cdot6}\times\frac{3\cdot 2}{4\cdot1}f_0. \end{align} But $$\displaystyle \frac{1}{f_0}=\frac{\Gamma\left(\frac{1}{5}\right)\Gamma\left(\frac{4}{5}\right)}{\Gamma\left(\frac{2}{5}\right)\Gamma\left(\frac{3}{5}\right)}=\frac{\sin\frac{2\pi}{5}}{\sin\frac{\pi}{5}}=\frac{1+\sqrt5}{2}$$ and $f_{N\rightarrow\infty}=1$ since $$\lim_{x\rightarrow +\infty}\frac{\Gamma(x-a)\Gamma(x+a)}{\Gamma(x-b)\Gamma(x+b)}=1.$$ The infinite fraction representation follows. $\blacksquare$