Question Number 28622 by abdo imad last updated on 28/Jan/18
$${find}\:\:\sum_{{k}=\mathrm{0}} ^{+\infty} \:{arctan}\left(\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} +{k}+\mathrm{1}}\right)\:. \\ $$
Commented by abdo imad last updated on 30/Jan/18
$${let}\:{put}\:{S}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{arctan}\:\left(\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} +{k}+\mathrm{1}}\right)\:{we}\:{have} \\ $$$$\:\frac{\mathrm{1}}{{k}^{\mathrm{2}} +{k}+\mathrm{1}}=\:\:\frac{\mathrm{1}}{\mathrm{1}+{k}\left({k}+\mathrm{1}\right)}=\:\frac{{k}+\mathrm{1}−{k}}{\mathrm{1}+{k}\left({k}+\mathrm{1}\right)\:}\:{let}\:{put}\:\:{k}={tan}\left({u}_{{k}} \right) \\ $$$$=\:\frac{{tan}\left({u}_{{k}+\mathrm{1}} \right)\:−{tan}\left({u}_{{k}} \right)}{\mathrm{1}+{tan}\left({u}_{{k}} \right){tan}\left({u}_{{k}+\mathrm{1}} \right)}={tan}\left(\:{u}_{{k}+\mathrm{1}} \:−{u}_{{k}} \right)\Rightarrow \\ $$$${artan}\left(\frac{\mathrm{1}}{{k}^{\mathrm{2}} +{k}+\mathrm{1}}\right)={u}_{{k}+\mathrm{1}} \:−{u}_{{k}} \:\:\:\:{and} \\ $$$${S}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(\:{u}_{{k}+\mathrm{1}} −{u}_{{k}} \right)=\:{u}_{\mathrm{1}} −{u}_{\mathrm{0}} \:+{u}_{\mathrm{2}} −{u}_{\mathrm{1}\:} +....+{u}_{{n}+\mathrm{1}} −{u}_{{n}} \\ $$$$=\:{u}_{{n}+\mathrm{1}} \:−{u}_{\mathrm{0}} =\:{arctan}\left({n}+\mathrm{1}\right)\:\Rightarrow{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\frac{\pi}{\mathrm{4}}\:. \\ $$
Commented by abdo imad last updated on 30/Jan/18
$${lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} \:\:\:=\:\frac{\pi}{\mathrm{2}}. \\ $$