Question Number 27663 by abdo imad last updated on 12/Jan/18 | ||
$${let}\:{give}\:\:{U}_{{n}} ={n}\:\left(\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\:+\:\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} +{n}^{\mathrm{2}} }+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} +{n}^{\mathrm{2}} }\:+....\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} +{n}^{\mathrm{2}} }\right) \\ $$ $${find}\:{lim}_{{n}−>\propto} \:\:{U}_{{n}} \:\:\:. \\ $$ $$ \\ $$ | ||
Commented byabdo imad last updated on 13/Jan/18 | ||
$${we}\:{have}\:{U}_{{n}} =\:{n}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} \:+{n}^{\mathrm{2}} }=\:\frac{\mathrm{1}}{{n}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{{k}}{{n}}\right)^{\mathrm{2}} } \\ $$ $${U}_{{n}} \:\:{is}\:{aRieman}\:{sum}\:{and} \\ $$ $${lim}_{{n}−>\propto} {U}_{{n}} \:=\:{lim}_{{n}−>\propto\:} \:\:\frac{\mathrm{1}−\mathrm{0}}{{n}}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{{k}\left(\mathrm{1}−\mathrm{0}\right)}{{n}}\right)^{\mathrm{2}} } \\ $$ $$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:^{} =\left[{arctanx}\right]_{\mathrm{0}} ^{\mathrm{1}} =\:\frac{\pi}{\mathrm{4}}\:. \\ $$ | ||