Question-196429

Question Number 196429 by SANOGO last updated on 24/Aug/23

Answered by Frix last updated on 24/Aug/23

ln (Π_(k=1) ^∞ ((e)^(1/n) ((2/π))^(1/k^2 ) )) =Σ_(k=1) ^∞ (((ln 2 −ln π)/k^2 )+(1/n)) = =1+ln (2/π) Σ_(k=1) ^∞ (1/k^2 ) =1+(π^2 /6)ln (2/π) e^(1+(π^2 /6)ln (2/π)) =((2/π))^(π^2 /6) e

$$\mathrm{ln}\:\left(\underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\:\left(\sqrt[{{n}}]{\mathrm{e}}\left(\frac{\mathrm{2}}{\pi}\right)^{\frac{\mathrm{1}}{{k}^{\mathrm{2}} }} \right)\right)\:=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\frac{\mathrm{ln}\:\mathrm{2}\:−\mathrm{ln}\:\pi}{{k}^{\mathrm{2}} }+\frac{\mathrm{1}}{{n}}\right)\:= \\ $$$$=\mathrm{1}+\mathrm{ln}\:\frac{\mathrm{2}}{\pi}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:=\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\mathrm{ln}\:\frac{\mathrm{2}}{\pi} \\ $$$$\mathrm{e}^{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\mathrm{ln}\:\frac{\mathrm{2}}{\pi}} =\left(\frac{\mathrm{2}}{\pi}\right)^{\frac{\pi^{\mathrm{2}} }{\mathrm{6}}} \mathrm{e} \\ $$

Commented by SANOGO last updated on 24/Aug/23

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

Commented by Frix last updated on 24/Aug/23

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Question-196429

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