Question Number 219657 by Nicholas666 last updated on 30/Apr/25
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{prove}; \\ $$$$\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\left(\mathrm{5}{n}−\mathrm{2}\right)\left(\mathrm{5}{n}−\mathrm{3}\right)}{\left(\mathrm{5}{n}−\mathrm{1}\right)\left(\mathrm{5}{n}−\mathrm{4}\right)}\:=\:\varphi \\ $$$$ \\ $$
Answered by MrGaster last updated on 30/Apr/25
$$\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\left(\mathrm{5}{n}−\mathrm{2}\right)\left(\mathrm{5}{n}−\mathrm{3}\right)}{\left(\mathrm{5}{n}−\mathrm{1}\right)\left(\mathrm{5}{n}−\mathrm{4}\right)}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{4}}{\mathrm{5}}\right)}{\Gamma\left(\frac{\mathrm{2}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{5}}\right)} \\ $$$$\Gamma\left({z}\right)\Gamma\left(\mathrm{1}−{z}\right)=\frac{\pi}{\mathrm{sin}\left(\pi{z}\right)}\Rightarrow\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{4}}{\mathrm{5}}\right)}{\Gamma\left(\frac{\mathrm{2}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{5}}\right)}= \\ $$$$\frac{\mathrm{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)}{\mathrm{sin}\left(\frac{\pi}{\mathrm{5}}\right)}=\mathrm{2}\:\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}\right)=\varphi \\ $$