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Question Number 173721 by KONE last updated on 16/Jul/22

Answered by mnjuly1970 last updated on 17/Jul/22

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

Commented by KONE last updated on 17/Jul/22

$$mercibienavous$$

Commented by mnjuly1970 last updated on 18/Jul/22

$$youarewelcom$$

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