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Question Number 144607 by aliibrahim1 last updated on 26/Jun/21

Answered by mathmax by abdo last updated on 27/Jun/21

A_n =Π_(k=2) ^n e(1−(1/k^2 ))^k^2 ⇒Π_(n=2) ^∞ (....)=lim_(n⌣∞) A_n A_n =e^(n−1) Π_(k=2) ^n (1−(1/k^2 ))^k^2 ⇒log(A_n )=n−1+Σ_(k=2) ^n k^2 log(1−(1/k^2 )) log^′ (1−x)=−(1/(1−x))=−(1+x+x^2 +o(x^3 )) ⇒ log(1−x)∼−x−(x^2 /2) ⇒log(1−(1/k^2 ))∼−(1/k^2 )−(1/(2k^4 )) ⇒ k^2 log(1−(1/k^2 ))∼−1−(1/(2k^2 )) ⇒Σ_(k=2) ^n k^2 log(...)∼−(n−1) −(1/2)Σ_(k=2) ^n (1/k^2 ) ⇒log(A_n )∼−(1/2)Σ_(k=2) ^n (1/k^2 ) =−(1/2)(ξ_n (2)−1) =(1/2)−((ξ_n (2))/2) ⇒ lim_(n→+∞) log(A_n )=(1/2)−(π^2 /(12)) ⇒lim_(n→+∞) A_n =e^((1/2)−(π^2 /(12)))

$$\mathrm{A}_{\mathrm{n}} =\prod_{\mathrm{k}=\mathrm{2}} ^{\mathrm{n}} \mathrm{e}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\right)^{\mathrm{k}^{\mathrm{2}} } \:\Rightarrow\prod_{\mathrm{n}=\mathrm{2}} ^{\infty} \left(....\right)=\mathrm{lim}_{\mathrm{n}\smile\infty} \mathrm{A}_{\mathrm{n}} \\ $$$$\mathrm{A}_{\mathrm{n}} =\mathrm{e}^{\mathrm{n}−\mathrm{1}} \:\prod_{\mathrm{k}=\mathrm{2}} ^{\mathrm{n}} \:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\right)^{\mathrm{k}^{\mathrm{2}} } \:\Rightarrow\mathrm{log}\left(\mathrm{A}_{\mathrm{n}} \right)=\mathrm{n}−\mathrm{1}+\sum_{\mathrm{k}=\mathrm{2}} ^{\mathrm{n}} \mathrm{k}^{\mathrm{2}} \:\mathrm{log}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\right) \\ $$$$\mathrm{log}^{'} \left(\mathrm{1}−\mathrm{x}\right)=−\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}=−\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} +\mathrm{o}\left(\mathrm{x}^{\mathrm{3}} \right)\right)\:\Rightarrow \\ $$$$\mathrm{log}\left(\mathrm{1}−\mathrm{x}\right)\sim−\mathrm{x}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{log}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\right)\sim−\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2k}^{\mathrm{4}} }\:\Rightarrow \\ $$$$\mathrm{k}^{\mathrm{2}} \mathrm{log}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\right)\sim−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2k}^{\mathrm{2}} }\:\Rightarrow\sum_{\mathrm{k}=\mathrm{2}} ^{\mathrm{n}} \mathrm{k}^{\mathrm{2}} \mathrm{log}\left(...\right)\sim−\left(\mathrm{n}−\mathrm{1}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{k}=\mathrm{2}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} }\:\Rightarrow\mathrm{log}\left(\mathrm{A}_{\mathrm{n}} \right)\sim−\frac{\mathrm{1}}{\mathrm{2}}\sum_{\mathrm{k}=\mathrm{2}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}^{\mathrm{2}} } \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left(\xi_{\mathrm{n}} \left(\mathrm{2}\right)−\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\xi_{\mathrm{n}} \left(\mathrm{2}\right)}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{log}\left(\mathrm{A}_{\mathrm{n}} \right)=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:\Rightarrow\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{A}_{\mathrm{n}} =\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}} \\ $$$$ \\ $$

Commented by aliibrahim1 last updated on 27/Jun/21

thx sir

$${thx}\:{sir} \\ $$

Commented by mathmax by abdo last updated on 28/Jun/21

you are welcome

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

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