Question Number 103881 by bemath last updated on 18/Jul/20

Π_(n=1) ^∞ (((2n−1)(2n+1))/(4n^2 )) ?

Answered by bramlex last updated on 18/Jul/20

use Euler infinity product   ((sin x)/x) = Π_(n=1) ^∞ (1−(x^2 /(n^2 π^2 )))  put x = (π/2) ⇒((sin ((π/2)))/(π/2)) = Π_(n=1) ^∞ (1−((π^2 /4)/(n^2 π^2 )))  (2/π) = Π_(n=1) ^∞ (1−(1/(4n^2 )))   (2/π) = Π_(n=1) ^∞ (((4n^2 −1)/(4n^2 )))= Π_(n=1) ^∞ (((2n−1)(2n+1))/(4n^2 )).

Commented bybemath last updated on 18/Jul/20

cooll

Answered by mathmax by abdo last updated on 18/Jul/20

we have Π_(n=1) ^∞  (((2n−1)(2n+1))/(4n^2 )) =Π_(n=1) ^∞  ((4n^2 −1)/(4n^2 )) =Π_(n=1) ^∞ (1−(1/(4n^2 )))  we know sinz =z Π_(n=1) ^∞ (1−(z^2 /(n^2 π^2 )))  we choose z /(z^2 /π^2 ) =(1/4) ⇒z^2  =(π^2 /4) ⇒  z =(π/2) ⇒sin((π/2)) =(π/2) Π_(n=1) ^∞ (1−(1/(4n^2 ))) ⇒Π_(n=1) ^∞ (1−(1/(4n^2 ))) =(2/π)

Commented bybemath last updated on 18/Jul/20

thank you sir