Question Number 30235 by NECx last updated on 18/Feb/18
$${find}\:{the}\:{sum}\:{of}\:{the}\:{infinite} \\ $$$${series}\: \\ $$$$\:\:\:\:\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\right) \\ $$
Commented by prof Abdo imad last updated on 18/Feb/18
$${for}\:{n}\geqslant\mathrm{1}\:\:\:\frac{\mathrm{2}}{{n}^{\mathrm{2}} }=\:\frac{{n}+\mathrm{1}−\left({n}−\mathrm{1}\right)}{\mathrm{1}+\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)}\:{let}\:{put}\:{n}={tanu}_{{n}} \\ $$$$\frac{\mathrm{2}}{{n}^{\mathrm{2}} }=\:\frac{{tanu}_{{n}+\mathrm{1}} −{tan}_{{n}−\mathrm{1}} }{\mathrm{1}+{tanu}_{{n}} {tanu}_{{n}−\mathrm{1}} }={tan}\left({u}_{{n}+\mathrm{1}} −{u}_{{n}−\mathrm{1}} \right) \\ $$$${arctan}\left(\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\right)=\:{u}_{{n}+\mathrm{1}} \:−{u}_{{n}−\mathrm{1}\:} \:{and} \\ $$$${S}_{{N}} =\sum_{{n}=\mathrm{1}} ^{{N}} {arctan}\left(\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\right)=\sum_{{n}=\mathrm{1}} ^{{N}} \left(\left({u}_{{n}+\mathrm{1}} −{u}_{{n}} \right)\:+\left({u}_{{n}} −{u}_{{n}−\mathrm{1}} \right)\right) \\ $$$$=\sum_{{n}=\mathrm{1}} ^{{N}} \left({u}_{{n}+\mathrm{1}} \:−{u}_{{n}} \right)\:+\sum_{{n}=\mathrm{1}} ^{{N}} \left({u}_{{n}} \:−{u}_{{n}−\mathrm{1}} \right) \\ $$$$={u}_{{N}+\mathrm{1}} \:−{u}_{\mathrm{1}} \:\:+{u}_{{N}} \:−{u}_{\mathrm{0}\:} ={arctan}\left({N}+\mathrm{1}\right)+{arctanN}\:−\frac{\pi}{\mathrm{4}} \\ $$$${lim}_{{N}\rightarrow\infty} \:{S}_{{N}} =\:\frac{\pi}{\mathrm{2}}\:+\frac{\pi}{\mathrm{2}}\:−\frac{\pi}{\mathrm{4}}=\:\frac{\mathrm{3}\pi}{\mathrm{4}}\:. \\ $$
Commented by abdo imad last updated on 18/Feb/18
$${arctan}\:{means}\:{tan}^{−\mathrm{1}} . \\ $$
Commented by NECx last updated on 19/Feb/18
$${thank}\:{you}\:{so}\:{much} \\ $$