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Question Number 63919 by raj last updated on 11/Jul/19

Π_(n=2) ^∞ (1−(1/n^2 ))=?

$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)=? \\ $$

Commented by Prithwish sen last updated on 11/Jul/19

Π_(n=2) ^(n=n) (1−(1/n^2 )) = ((1.2.3^2 .4^2 ...........n.(n+1))/(1^2 .2^2 .3^2 ..........(n−1)^2 .n^2 )) = [(((n+1)!)/(n!))]^2 .(1/(2n(n+1))) = (((n+1)^2 )/(2n(n+1))) = (1/2)(1+(1/n)) → (1/2) as n→∞ please check.

$$\underset{\mathrm{n}=\mathrm{2}} {\overset{\mathrm{n}=\boldsymbol{\mathrm{n}}} {\prod}}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\right)\:=\:\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}^{\mathrm{2}} .\mathrm{4}^{\mathrm{2}} ...........\mathrm{n}.\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{1}^{\mathrm{2}} .\mathrm{2}^{\mathrm{2}} .\mathrm{3}^{\mathrm{2}} ..........\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} .\mathrm{n}^{\mathrm{2}} }\: \\ $$$$=\:\left[\frac{\left(\mathrm{n}+\mathrm{1}\right)!}{\mathrm{n}!}\right]^{\mathrm{2}} .\frac{\mathrm{1}}{\mathrm{2n}\left(\mathrm{n}+\mathrm{1}\right)}\:=\:\frac{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2n}\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}\right)\:\rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{as}\:\mathrm{n}\rightarrow\infty \\ $$$$\mathrm{please}\:\mathrm{check}. \\ $$

Commented by mathmax by abdo last updated on 11/Jul/19

let A_n =Π_(k=2) ^n (1−(1/k^2 )) ⇒Π_(k=2) ^∞ (1−(1/k^2 )) =lim_(n→+∞) A_n but A_n =Π_(k=2) ^n ((k^2 −1)/k^2 ) =Π_(k=2) ^n ((k−1)/k) ((k+1)/k) =Π_(k=2) ^n ((k−1)/k) Π_(k=2) ^n ((k+1)/k) =(1/2) (2/3)...((n−2)/(n−1))((n−1)/n) ×(3/2).(4/3).....(n/(n−1)) ((n+1)/n) =(1/n) ((n+1)/2) =((n+1)/(2n)) ⇒lim_(n→+∞) A_n =(1/2)

$${let}\:{A}_{{n}} =\prod_{{k}=\mathrm{2}} ^{{n}} \left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)\:\Rightarrow\prod_{{k}=\mathrm{2}} ^{\infty} \:\left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)\:={lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$$${but}\:{A}_{{n}} =\prod_{{k}=\mathrm{2}} ^{{n}} \:\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} }\:=\prod_{{k}=\mathrm{2}} ^{{n}} \:\frac{{k}−\mathrm{1}}{{k}}\:\frac{{k}+\mathrm{1}}{{k}} \\ $$$$=\prod_{{k}=\mathrm{2}} ^{{n}} \:\frac{{k}−\mathrm{1}}{{k}}\:\prod_{{k}=\mathrm{2}} ^{{n}} \:\frac{{k}+\mathrm{1}}{{k}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{2}}{\mathrm{3}}...\frac{{n}−\mathrm{2}}{{n}−\mathrm{1}}\frac{{n}−\mathrm{1}}{{n}}\:×\frac{\mathrm{3}}{\mathrm{2}}.\frac{\mathrm{4}}{\mathrm{3}}.....\frac{{n}}{{n}−\mathrm{1}}\:\frac{{n}+\mathrm{1}}{{n}} \\ $$$$=\frac{\mathrm{1}}{{n}}\:\frac{{n}+\mathrm{1}}{\mathrm{2}}\:=\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by raj last updated on 11/Jul/19

thank you

$${thank}\:{you} \\ $$

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