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Question Number 151673 by mathdanisur last updated on 22/Aug/21

Answered by Kamel last updated on 22/Aug/21

$$L=\underset{n\to +\mathrm{\infty }}{lim}\underset{k=n}{\prod ^{2n}}\frac{\pi }{\pi -Arctan\left(\frac{1}{k}\right)}=\underset{n\to +\mathrm{\infty }}{lim}\underset{k=n}{\prod ^{2n}}\frac{\pi k}{\pi k-1}$$$$=\underset{n\to +\mathrm{\infty }}{lim}\frac{n.(n+1)(n+2)...\left(2n\right)}{(n-\frac{1}{\pi })(n+1-\frac{1}{\pi })...(2n-\frac{1}{\pi })}$$$$=\underset{n\to +\mathrm{\infty }}{lim}\frac{\mathrm{\Gamma }(2n+1)\mathrm{\Gamma }(n-\frac{1}{\pi })}{\mathrm{\Gamma }(2n-\frac{1}{\pi }+1)\mathrm{\Gamma }\left(n\right)}=\underset{n\to +\mathrm{\infty }}{lim}\frac{{\left(n\right)}^{2n}{(n-\frac{1}{\pi }-1)}^n{4}^{\frac{1}{\pi }}}{{(n-1)}^n{(n-\frac{1}{2\pi })}^{2n}}$$$$=\underset{n\to +\mathrm{\infty }}{lim}\frac{{(1-\frac{1}{\pi (n-1)})}^{n-1}{4}^{\frac{1}{\pi }}}{{(1-\frac{1}{2\pi n})}^{2n}}=\sqrt[\pi ]{4}$$$$$$

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