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Question Number 219657 by Nicholas666 last updated on 30/Apr/25

prove; Π_(n=1) ^∞ (((5n−2)(5n−3))/((5n−1)(5n−4))) = ϕ

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{prove}; \\ $$$$\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\left(\mathrm{5}{n}−\mathrm{2}\right)\left(\mathrm{5}{n}−\mathrm{3}\right)}{\left(\mathrm{5}{n}−\mathrm{1}\right)\left(\mathrm{5}{n}−\mathrm{4}\right)}\:=\:\varphi \\ $$$$ \\ $$

Answered by MrGaster last updated on 30/Apr/25

Π_(n=1) ^∞ (((5n−2)(5n−3))/((5n−1)(5n−4)))=((Γ((1/5))Γ((4/5)))/(Γ((2/5))Γ((3/5)))) Γ(z)Γ(1−z)=(π/(sin(πz)))⇒((Γ((1/5))Γ((4/5)))/(Γ((2/5))Γ((3/5))))= ((sin(((2π)/5)))/(sin((π/5))))=2 cos((π/5))=ϕ

$$\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\left(\mathrm{5}{n}−\mathrm{2}\right)\left(\mathrm{5}{n}−\mathrm{3}\right)}{\left(\mathrm{5}{n}−\mathrm{1}\right)\left(\mathrm{5}{n}−\mathrm{4}\right)}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{4}}{\mathrm{5}}\right)}{\Gamma\left(\frac{\mathrm{2}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{5}}\right)} \\ $$$$\Gamma\left({z}\right)\Gamma\left(\mathrm{1}−{z}\right)=\frac{\pi}{\mathrm{sin}\left(\pi{z}\right)}\Rightarrow\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{4}}{\mathrm{5}}\right)}{\Gamma\left(\frac{\mathrm{2}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{5}}\right)}= \\ $$$$\frac{\mathrm{sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)}{\mathrm{sin}\left(\frac{\pi}{\mathrm{5}}\right)}=\mathrm{2}\:\mathrm{cos}\left(\frac{\pi}{\mathrm{5}}\right)=\varphi \\ $$

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