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

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