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Question Number 141378 by mnjuly1970 last updated on 18/May/21

$$provethat::$$$$\underset{n=0}{\prod ^{\mathrm{\infty }}}\frac{(5n+2)(5n+3)}{(5n+1)(5n+4)}=\phi$$$$\phi :=\frac{1+\sqrt{5}}{2}$$

Answered by Dwaipayan Shikari last updated on 18/May/21

$$\underset{n=0}{\prod ^{\mathrm{\infty }}}\frac{(5n+2)(5n+3)}{(5n+1)(5n+4)}=\underset{n=0}{\prod ^{\mathrm{\infty }}}\frac{(n+\frac{2}{5})(n+\frac{3}{5})}{(n+\frac{1}{5})(n+\frac{4}{5})}$$$$\underset{n=0}{\prod ^N}\frac{(n+\frac{2}{5})(n+\frac{3}{5})}{(n+\frac{1}{5})(n+\frac{4}{5})}=\frac{{\left(\frac{2}{5}\right)}_N{\left(\frac{3}{5}\right)}_N}{{\left(\frac{1}{5}\right)}_N{\left(\frac{4}{5}\right)}_N}{\left(a\right)}_n=\frac{\mathrm{\Gamma }(n+a)}{\mathrm{\Gamma }\left(a\right)}$$$$=\frac{\mathrm{\Gamma }(N+\frac{2}{5})\mathrm{\Gamma }(N+\frac{3}{5})}{\mathrm{\Gamma }(N+\frac{1}{5})\mathrm{\Gamma }(N+\frac{4}{5})}.\frac{\mathrm{\Gamma }\left(\frac{1}{5}\right)\mathrm{\Gamma }\left(\frac{4}{5}\right)}{\mathrm{\Gamma }\left(\frac{2}{5}\right)\mathrm{\Gamma }\left(\frac{3}{5}\right)}=\frac{\mathrm{\Gamma }(N+\frac{2}{5})\mathrm{\Gamma }(N+\frac{3}{5})}{\mathrm{\Gamma }(N+\frac{1}{5})\mathrm{\Gamma }(N+\frac{4}{5})}.\frac{\pi sin\left(\frac{2}{5}\pi \right)}{\pi sin\left(\frac{\pi }{5}\right)}$$$$WhenN\to \mathrm{\infty }itis\frac{{(N-\frac{3}{5})}^{N-\frac{3}{5}}{e}^{-N+\frac{3}{5}}(N-\frac{2}{5})e^{-N+\frac{2}{5}}}{{(N-\frac{4}{5})}^{N-\frac{4}{5}}(N-\frac{1}{5})e^{-2N+1}}.\frac{sin\left(\frac{2\pi }{5}\right)}{sin\left(\frac{\pi }{5}\right)}$$$$=1.\frac{2sin\left(\frac{\pi }{5}\right)cos\left(\frac{\pi }{5}\right)}{sin\left(\frac{\pi }{5}\right)}=2cos\left(\frac{\pi }{5}\right)=2.\frac{\sqrt{5}+1}{4}=\frac{\sqrt{5}+1}{2}=\phi$$

Commented by mnjuly1970 last updated on 18/May/21

$$thanksalotmrpayan...$$

Answered by bramlexs22 last updated on 18/May/21

$$P=\underset{n=0}{\prod ^{\mathrm{\infty }}}\frac{(n+\frac{2}{5})(n+\frac{3}{5})}{(n+\frac{1}{5})(n+\frac{4}{5})}=$$$$\frac{\mathrm{\Gamma }\left(\frac{1}{5}\right)\mathrm{\Gamma }\left(\frac{4}{5}\right)}{\mathrm{\Gamma }\left(\frac{2}{5}\right)\mathrm{\Gamma }\left(\frac{3}{5}\right)}=\frac{\sin \left(\frac{2\pi }{5}\right)}{\sin \left(\frac{\pi }{5}\right)}=2\cos \left(\frac{\pi }{5}\right)$$$$=2\left(\frac{\sqrt{5}+1}{4}\right)=\frac{\sqrt{5}+1}{2}$$

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