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))) =?

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...