Let be P the set of prime numbers and A=P∪{0,1} Prove that Π_(n∉A) (n/( (√(n^2 −1)))) =(2/π)(√3)


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Question Number 117838 by snipers237 last updated on 13/Oct/20
$Let be P the set of prime numbers and A=P∪{0,1} Prove that Π_(n∉A) (n/( (√(n^2 −1)))) =(2/π)(√3)$
$L e t b e P t h e s e t o f p r i m e n u m b e r s a n d$ $A = P \cup {0, 1}$ $P r o v e t h a t \prod_{n \notin A} \frac{n}{\sqrt{n^{2} - 1}} = \frac{2}{π} \sqrt{3}$
	Answered by mindispower last updated on 14/Oct/20
	$let Π_(n≥2) (n/( (√(n^2 −1))))=I I^2 =Π_(n≥2) (n/((n−1))).(n/((n+1)))=((2.2)/(1.3)).((3.3)/(2.4)).((4.4)/(3.5))...... I^2 =lim_(N→∞) .Π_(n=2) ^N ((n/(n−1)).(n/(n+1)))=lim_(N→∞) ((2N)/(N+1)) I^2 →2⇒I→(√2),since I≥0 and x→(√(x )) continus let P bee set ofrimes numbers Π_(p∈P) (n^2 /(n^2 −1))=Π_(p∈P) (1/(1−(1/p^2 ))) =Π_(p∈P) (Σ_(s≥0) (1/p^(2s) ))=Σ_(n≥1) (1/n^2 )=ζ(2)=(π^2 /6) A=P∪{0,1} Π_(n∉A) (n^2 /(n^2 −1))=Π_(n≥2) (n^2 /(n^2 −1))/(Π_(n∈P) (n^2 /(n^2 −1)))=(2/(π^2 /6)) =((12)/π^2 ) Π_(n∉A) (n/( (√(n^2 −1))))=(√((12)/π^2 ))=((2(√3))/π)$
	$l e t \prod_{n ⩾ 2} \frac{n}{\sqrt{n^{2} - 1}} = I$ $I^{2} = \prod_{n ⩾ 2} \frac{n}{(n - 1)} . \frac{n}{(n + 1)} = \frac{2.2}{1.3} . \frac{3.3}{2.4} . \frac{4.4}{3.5} . . . . . .$ $I^{2} = \lim_{N \to \infty} . \underset{n = 2}{\prod^{N}} (\frac{n}{n - 1} . \frac{n}{n + 1}) = \lim_{N \to \infty} \frac{2 N}{N + 1}$ $I^{2} \to 2 \Rightarrow I \to \sqrt{2}, s i n c e I ⩾ 0 a n d x \to \sqrt{x} c o n t i n u s$ $l e t P b e e s e t o f r i m e s n u m b e r s$ $\prod_{p \in P} \frac{n^{2}}{n^{2} - 1} = \prod_{p \in P} \frac{1}{1 - \frac{1}{p^{2}}}$ $= \prod_{p \in P} (\sum_{s ⩾ 0} \frac{1}{p^{2 s}}) = \sum_{n ⩾ 1} \frac{1}{n^{2}} = ζ (2) = \frac{π^{2}}{6}$ $A = P \cup {0, 1}$ $\prod_{n \notin A} \frac{n^{2}}{n^{2} - 1} = \prod_{n ⩾ 2} \frac{n^{2}}{n^{2} - 1} / (\prod_{n \in P} \frac{n^{2}}{n^{2} - 1}) = \frac{2}{\frac{π^{2}}{6}}$ $= \frac{12}{π^{2}}$ $\prod_{n \notin A} \frac{n}{\sqrt{n^{2} - 1}} = \sqrt{\frac{12}{π^{2}}} = \frac{2 \sqrt{3}}{π}$
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