Question Number 141378 by mnjuly1970 last updated on 18/May/21
$$\:\:\:{prove}\:{that}:: \\ $$$$\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\frac{\left(\mathrm{5}{n}+\mathrm{2}\right)\left(\mathrm{5}{n}+\mathrm{3}\right)}{\left(\mathrm{5}{n}+\mathrm{1}\right)\left(\mathrm{5}{n}+\mathrm{4}\right)}\:=\varphi\: \\ $$$$\:\:\:\:\:\:\:\varphi:=\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Answered by Dwaipayan Shikari last updated on 18/May/21
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\frac{\left(\mathrm{5}{n}+\mathrm{2}\right)\left(\mathrm{5}{n}+\mathrm{3}\right)}{\left(\mathrm{5}{n}+\mathrm{1}\right)\left(\mathrm{5}{n}+\mathrm{4}\right)}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\frac{\left({n}+\frac{\mathrm{2}}{\mathrm{5}}\right)\left({n}+\frac{\mathrm{3}}{\mathrm{5}}\right)}{\left({n}+\frac{\mathrm{1}}{\mathrm{5}}\right)\left({n}+\frac{\mathrm{4}}{\mathrm{5}}\right)} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{{N}} {\prod}}\frac{\left({n}+\frac{\mathrm{2}}{\mathrm{5}}\right)\left({n}+\frac{\mathrm{3}}{\mathrm{5}}\right)}{\left({n}+\frac{\mathrm{1}}{\mathrm{5}}\right)\left({n}+\frac{\mathrm{4}}{\mathrm{5}}\right)}=\frac{\left(\frac{\mathrm{2}}{\mathrm{5}}\right)_{{N}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right)_{{N}} }{\left(\frac{\mathrm{1}}{\mathrm{5}}\right)_{{N}} \left(\frac{\mathrm{4}}{\mathrm{5}}\right)_{{N}} }\:\:\:\left({a}\right)_{{n}} =\frac{\Gamma\left({n}+{a}\right)}{\Gamma\left({a}\right)} \\ $$$$=\frac{\Gamma\left({N}+\frac{\mathrm{2}}{\mathrm{5}}\right)\Gamma\left({N}+\frac{\mathrm{3}}{\mathrm{5}}\right)}{\Gamma\left({N}+\frac{\mathrm{1}}{\mathrm{5}}\right)\Gamma\left({N}+\frac{\mathrm{4}}{\mathrm{5}}\right)}.\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{4}}{\mathrm{5}}\right)}{\Gamma\left(\frac{\mathrm{2}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{5}}\right)}=\frac{\Gamma\left({N}+\frac{\mathrm{2}}{\mathrm{5}}\right)\Gamma\left({N}+\frac{\mathrm{3}}{\mathrm{5}}\right)}{\Gamma\left({N}+\frac{\mathrm{1}}{\mathrm{5}}\right)\Gamma\left({N}+\frac{\mathrm{4}}{\mathrm{5}}\right)}.\frac{\pi{sin}\left(\frac{\mathrm{2}}{\mathrm{5}}\pi\right)}{\pi{sin}\left(\frac{\pi}{\mathrm{5}}\right)} \\ $$$${When}\:{N}\rightarrow\infty\:\:{it}\:{is}\:\frac{\left({N}−\frac{\mathrm{3}}{\mathrm{5}}\right)^{{N}−\frac{\mathrm{3}}{\mathrm{5}}} {e}^{−{N}+\frac{\mathrm{3}}{\mathrm{5}}} \left({N}−\frac{\mathrm{2}}{\mathrm{5}}\right){e}^{−{N}+\frac{\mathrm{2}}{\mathrm{5}}} }{\left({N}−\frac{\mathrm{4}}{\mathrm{5}}\right)^{{N}−\frac{\mathrm{4}}{\mathrm{5}}} \left({N}−\frac{\mathrm{1}}{\mathrm{5}}\right){e}^{−\mathrm{2}{N}+\mathrm{1}} }.\frac{{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)}{{sin}\left(\frac{\pi}{\mathrm{5}}\right)} \\ $$$$=\mathrm{1}.\frac{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{5}}\right){cos}\left(\frac{\pi}{\mathrm{5}}\right)}{{sin}\left(\frac{\pi}{\mathrm{5}}\right)}=\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{5}}\right)=\mathrm{2}.\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}=\varphi \\ $$
Commented by mnjuly1970 last updated on 18/May/21
$${thanks}\:{alot}\:{mr}\:{payan}… \\ $$
Answered by bramlexs22 last updated on 18/May/21
$${P}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\:\frac{\left({n}+\frac{\mathrm{2}}{\mathrm{5}}\right)\left({n}+\frac{\mathrm{3}}{\mathrm{5}}\right)}{\left({n}+\frac{\mathrm{1}}{\mathrm{5}}\right)\left({n}+\frac{\mathrm{4}}{\mathrm{5}}\right)}\:= \\ $$$$\:\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{4}}{\mathrm{5}}\right)}{\Gamma\left(\frac{\mathrm{2}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{5}}\right)}\:=\:\frac{\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{5}}\right)}\:=\:\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{5}}\right) \\ $$$$\:=\:\mathrm{2}\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}\right)\:=\:\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$