Question Number 65700 by Masumsiddiqui399@gmail.com last updated on 02/Aug/19
Commented by Prithwish sen last updated on 02/Aug/19
$$=\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\left[\mathrm{1}+\left(\frac{\mathrm{k}}{\mathrm{n}}\right)^{\mathrm{2}} \right]}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{tan}^{−\mathrm{1}} \mathrm{x}\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\pi}{\mathrm{4}} \\ $$