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Question Number 65700 by Masumsiddiqui399@gmail.com last updated on 02/Aug/19

Commented by Prithwish sen last updated on 02/Aug/19

$$={\lim }_{\mathrm{n}\to \mathrm{\infty }}\frac{1}{\mathrm{n}}\underset{\mathrm{k}=0}{\sum ^1}\frac{1}{[1+{\left(\frac{\mathrm{k}}{\mathrm{n}}\right)}^2]}={\int }_0^1\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}={\tan }^{-1}\mathrm{x}{\mid }_0^1$$$$=\frac{\pi }{4}$$

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