Question Number 46598 by maxmathsup by imad last updated on 29/Oct/18
$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{arctan}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+{n}+\mathrm{1}}\right) \\ $$
Commented by math khazana by abdo last updated on 30/Oct/18
$${let}\:{u}_{{n}} =\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}\:+\mathrm{1}}\:\Rightarrow{u}_{{n}} =\frac{{n}+\mathrm{1}−{n}}{\mathrm{1}+\left({n}+\mathrm{1}\right){n}} \\ $$$${let}\:{n}={tanv}_{{n}} \:\Rightarrow{u}_{{n}} =\frac{{tanv}_{{n}+\mathrm{1}} −{tanv}_{{n}} }{\mathrm{1}+{v}_{{n}} .{v}_{{n}+\mathrm{1}} } \\ $$$$={tan}\left({v}_{{n}+\mathrm{1}} −{v}_{{n}} \right)\:\Rightarrow{arctan}\left({u}_{{n}} \right)={v}_{{n}+\mathrm{1}} −{v}_{{n}} \\ $$$$\Rightarrow\sum_{{k}=\mathrm{0}} ^{{n}} \:{arctan}\left({u}_{{k}} \right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:\left({v}_{{k}+\mathrm{1}} −{v}_{{k}} \right) \\ $$$$={v}_{{n}+\mathrm{1}} −{v}_{\mathrm{0}} ={arctan}\left({n}\right)\rightarrow\frac{\pi}{\mathrm{2}}\:\left({n}\rightarrow+\infty\right)\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:{arctan}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+{n}+\mathrm{1}}\right)=\frac{\pi}{\mathrm{2}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18
$${T}_{{n}} ={tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{n}+{n}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:={tan}^{−\mathrm{1}} \left\{\frac{{n}+\mathrm{1}−{n}}{\mathrm{1}+{n}\left({n}+\mathrm{1}\right)}\right\} \\ $$$$\:\:\:\:\:\:{T}_{{n}} ={tan}^{−\mathrm{1}} \left({n}+\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left({n}\right) \\ $$$$\:{T}_{\mathrm{1}} ={tan}^{−\mathrm{1}} \left(\mathrm{2}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$$\:\:{T}_{\mathrm{2}} ={tan}^{−\mathrm{1}} \left(\mathrm{3}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{2}\right) \\ $$$$\:\:{thus}\:{on}\:{addition} \\ $$$${S}_{{n}} ={T}_{\mathrm{1}} +{T}_{\mathrm{2}} +…+{T}_{{n}} \\ $$$${S}_{{n}} ={tan}^{−\mathrm{1}} \left({n}+\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$${now}\:{S}_{{n}} +{T}_{\mathrm{0}} ={tan}^{−\mathrm{1}} \left({n}+\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{1}\right)+{tan}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$${so}\:{value}\:{of} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}\right) \\ $$$$={tan}^{−\mathrm{1}} \left(\infty+\mathrm{1}\right)−{tan}^{−\mathrm{1}} \left(\mathrm{1}\right)+{tan}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$$=\frac{\pi}{\mathrm{2}} \\ $$