Question Number 103436 by bramlex last updated on 15/Jul/20

Π_(n = 3) ^∞ (1−(4/n^2 )) = ?

Answered by Worm_Tail last updated on 15/Jul/20

Π_3 ^(oo) (1−(4/n^2 ))=a   (1−(4/3^2 ))(1−(4/4^2 ))(1+(4/5^2 ))...=a    ln((1−(4/3^2 ))(1−(4/4^2 ))(1+(4/5^2 ))...)=ln(a)    ln(1−(4/3^2 ))+ln(1−(4/4^2 ))+ln(1+(4/5^2 ))...=ln(a)     Σ_(n=3) ^(oo) (ln(1−(4/n^2 )))=lna      Σ_(n=3) ^(oo) (ln((((n+2)(n−2))/n^2 )))=lna      Σ_(n=3) ^(oo) (ln(n+2)+ln(n−2)−2ln(n))=lna      Σ_(n=3) ^(oo) ln(n+2)+Σ_(n=3) ^(oo) ln(n−2)−2Σ_(n=3) ^(oo) ln(n)=lna      Σ_(n=3) ^(oo) ln(n+2)+(ln(1)+ln(2)+ln(3)+ln(4)+Σ_(n=7) ^(oo) ln(n−2))−2(ln(3)+ln(4)+Σ_(n=5) ^(oo) ln(n))=lna      Σ_(n=3) ^(oo) ln(n+2)=Σ_(n=7) ^(oo) ln(n−2)=Σ_(n=5) ^(oo) ln(n)=s     s+(ln(1)+ln(2)+ln(3)+ln(4)+s)−2(ln(3)+ln(4)+s)=lna      s+ln(1)+ln(2)+ln(3)+ln(4)+s−2ln(3)−2ln(4)−2s=lna      s+ln(1)+ln(2)+ln(3)+ln(4)+s−2ln(3)−2ln(4)−2s=lna   s+s−2s+ln(1)+ln(2)−ln(3)−ln(4)=ln(a)  ln((2/(12)))=ln(a)  (2/(12))=a  a=(1/6)

Commented bybobhans last updated on 15/Jul/20

waw..via logarithm

Answered by bemath last updated on 15/Jul/20

Commented bybobhans last updated on 15/Jul/20

cooll graphic