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Let-be-P-the-set-of-prime-numbers-and-A-P-0-1-Prove-that-n-A-n-n-2-1-2-pi-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)
$$\:\:{Let}\:{be}\:{P}\:\:{the}\:{set}\:{of}\:{prime}\:{numbers}\:{and}\: \\ $$$${A}={P}\cup\left\{\mathrm{0},\mathrm{1}\right\} \\ $$$${Prove}\:{that}\:\:\:\underset{{n}\notin{A}} {\prod}\:\frac{{n}}{\:\sqrt{{n}^{\mathrm{2}} −\mathrm{1}}}\:=\frac{\mathrm{2}}{\pi}\sqrt{\mathrm{3}}\: \\ $$
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)
$${let}\:\underset{{n}\geqslant\mathrm{2}} {\prod}\frac{{n}}{\:\sqrt{{n}^{\mathrm{2}} −\mathrm{1}}}={I} \\ $$$${I}^{\mathrm{2}} =\underset{{n}\geqslant\mathrm{2}} {\prod}\frac{{n}}{\left({n}−\mathrm{1}\right)}.\frac{{n}}{\left({n}+\mathrm{1}\right)}=\frac{\mathrm{2}.\mathrm{2}}{\mathrm{1}.\mathrm{3}}.\frac{\mathrm{3}.\mathrm{3}}{\mathrm{2}.\mathrm{4}}.\frac{\mathrm{4}.\mathrm{4}}{\mathrm{3}.\mathrm{5}}…… \\ $$$${I}^{\mathrm{2}} =\underset{{N}\rightarrow\infty} {\mathrm{lim}}.\underset{{n}=\mathrm{2}} {\overset{{N}} {\prod}}\left(\frac{{n}}{{n}−\mathrm{1}}.\frac{{n}}{{n}+\mathrm{1}}\right)=\underset{{N}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}{N}}{{N}+\mathrm{1}}\:\: \\ $$$${I}^{\mathrm{2}} \rightarrow\mathrm{2}\Rightarrow{I}\rightarrow\sqrt{\mathrm{2}},{since}\:{I}\geqslant\mathrm{0}\:{and}\:{x}\rightarrow\sqrt{{x}\:\:}\:{continus} \\ $$$${let}\:{P}\:{bee}\:{set}\:{ofrimes}\:{numbers} \\ $$$$\underset{{p}\in{P}} {\prod}\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{2}} −\mathrm{1}}=\underset{{p}\in{P}} {\prod}\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{{p}^{\mathrm{2}} }} \\ $$$$=\underset{{p}\in{P}} {\prod}\left(\underset{{s}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{p}^{\mathrm{2}{s}} }\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$${A}={P}\cup\left\{\mathrm{0},\mathrm{1}\right\} \\ $$$$\underset{{n}\notin{A}} {\prod}\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{2}} −\mathrm{1}}=\underset{{n}\geqslant\mathrm{2}} {\prod}\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{2}} −\mathrm{1}}/\left(\underset{{n}\in{P}} {\prod}\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{2}} −\mathrm{1}}\right)=\frac{\mathrm{2}}{\frac{\pi^{\mathrm{2}} }{\mathrm{6}}} \\ $$$$=\frac{\mathrm{12}}{\pi^{\mathrm{2}} } \\ $$$$\underset{{n}\notin{A}} {\prod}\frac{{n}}{\:\sqrt{{n}^{\mathrm{2}} −\mathrm{1}}}=\sqrt{\frac{\mathrm{12}}{\pi^{\mathrm{2}} }}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\pi} \\ $$

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