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Question Number 114758 by bobhans last updated on 21/Sep/20

Π_(n=1) ^∞ ((4n^2 (10n−6)(10n−4))/((2n−1)^2 (10n−1)(10n+1))) =?

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\mathrm{4}{n}^{\mathrm{2}} \left(\mathrm{10}{n}−\mathrm{6}\right)\left(\mathrm{10}{n}−\mathrm{4}\right)}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{10}{n}−\mathrm{1}\right)\left(\mathrm{10}{n}+\mathrm{1}\right)}\:=? \\ $$

Answered by Olaf last updated on 21/Sep/20

u_n = ((4n^2 (10n−6)(10n−4))/((2n−1)^2 (10n−1)(10n+1))) u_n = (((10n)^2 (10n−6)(10n−4))/((10n−5)^2 (10n−1)(10n+1))) u_n = (4/5)(((10n−0)(10n+0)(10n+4)(10n−4))/((10n+5)(10n−5)(10n−1)(10n+1))) Let Π = Π_(n=1) ^∞ u_n Π = (4/5)×((Π_(n=1) ^∞ (((10n+0)/(10n+1))×((10n+4)/(10n+5))))/(Π_(n=1) ^∞ (((10n−1)/(10n−0))×((10n−5)/(10n−4))))) work in progress...

$${u}_{{n}} \:=\:\frac{\mathrm{4}{n}^{\mathrm{2}} \left(\mathrm{10}{n}−\mathrm{6}\right)\left(\mathrm{10}{n}−\mathrm{4}\right)}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{10}{n}−\mathrm{1}\right)\left(\mathrm{10}{n}+\mathrm{1}\right)} \\ $$$${u}_{{n}} \:=\:\frac{\left(\mathrm{10}{n}\right)^{\mathrm{2}} \left(\mathrm{10}{n}−\mathrm{6}\right)\left(\mathrm{10}{n}−\mathrm{4}\right)}{\left(\mathrm{10}{n}−\mathrm{5}\right)^{\mathrm{2}} \left(\mathrm{10}{n}−\mathrm{1}\right)\left(\mathrm{10}{n}+\mathrm{1}\right)} \\ $$$${u}_{{n}} \:=\:\frac{\mathrm{4}}{\mathrm{5}}\frac{\left(\mathrm{10}{n}−\mathrm{0}\right)\left(\mathrm{10}{n}+\mathrm{0}\right)\left(\mathrm{10}{n}+\mathrm{4}\right)\left(\mathrm{10}{n}−\mathrm{4}\right)}{\left(\mathrm{10}{n}+\mathrm{5}\right)\left(\mathrm{10}{n}−\mathrm{5}\right)\left(\mathrm{10}{n}−\mathrm{1}\right)\left(\mathrm{10}{n}+\mathrm{1}\right)} \\ $$$$\mathrm{Let}\:\Pi\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}{u}_{{n}} \\ $$$$\Pi\:=\:\frac{\mathrm{4}}{\mathrm{5}}×\frac{\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\mathrm{10}{n}+\mathrm{0}}{\mathrm{10}{n}+\mathrm{1}}×\frac{\mathrm{10}{n}+\mathrm{4}}{\mathrm{10}{n}+\mathrm{5}}\right)}{\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\mathrm{10}{n}−\mathrm{1}}{\mathrm{10}{n}−\mathrm{0}}×\frac{\mathrm{10}{n}−\mathrm{5}}{\mathrm{10}{n}−\mathrm{4}}\right)} \\ $$$$\mathrm{work}\:\mathrm{in}\:\mathrm{progress}... \\ $$

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