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

Answered by Frix last updated on 24/Aug/23

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

Commented by SANOGO last updated on 24/Aug/23

$$thankyou$$

Commented by Frix last updated on 24/Aug/23

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