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Question Number 128065 by 0731619177 last updated on 04/Jan/21

Commented by mnjuly1970 last updated on 04/Jan/21

$${lim}_{n\to \mathrm{\infty }}\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{k^2}=\frac{{\pi }^2}{6}$$$${lim}_{n\to \mathrm{\infty }}(\underset{k=1}{\sum ^n}\frac{1}{k}-ln\left(n\right))=\gamma (euler-constant)$$$${lim}_{x\to \mathrm{\infty }}\left(\frac{\sqrt[x]{x!}}{x}\right)=\frac{1}{e}$$$$\therefore ans:{\left[{\left(\frac{1}{e}\right)}^{\gamma }\right]}^{\frac{{\pi }^2}{6}}={\left(\frac{1}{e}\right)}^{\frac{\gamma {\pi }^2}{6}}$$

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