Question Number 28682 by abdo imad last updated on 28/Jan/18
$${nature}\:{of}\:{the}\:{serie}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{tan}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{3}} }\right)\:. \\ $$
Commented by abdo imad last updated on 30/Jan/18
$${let}\:{put}\:{u}_{{n}} =\:\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{3}} }=\:\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:\Rightarrow\:{lim}_{{n}\rightarrow+\infty} \:{u}_{{n}\:} =\mathrm{0}\:{so} \\ $$$${tan}\left({u}_{{n}_{} } \right)\:\sim\:{u}_{{n}} \:\:\left({n}\:\rightarrow+\infty\right)\:{but}\:{the}\:{series}\:\sum_{{n}\geqslant\mathrm{1}} \:\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\:{and}\:\sum_{{n}\geqslant\mathrm{1}} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} } \\ $$$${converges}\:{so}\:{the}\:{serie}\:\sum_{{n}=\mathrm{0}} ^{\infty} {tan}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{3}} }\right)\:{converges}. \\ $$
Commented by abdo imad last updated on 30/Jan/18
$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{3}} }\right). \\ $$