Question Number 64671 by aliesam last updated on 20/Jul/19
Commented by mathmax by abdo last updated on 20/Jul/19
$${we}\:{have}\:\:\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:\leqslant\:{sinx}\:\leqslant\:{x}\:\:\:{for}\:{all}\:{x}\:{real}\:\Rightarrow \\ $$$${n}^{\mathrm{2}} {x}−\frac{{n}^{\mathrm{6}} \:{x}^{\mathrm{3}} }{\mathrm{6}}\:\leqslant{sin}\left({n}^{\mathrm{2}} {x}\right)\:\leqslant{n}^{\mathrm{2}} {x}\:\Rightarrow{nx}\:−\frac{{n}^{\mathrm{5}} }{\mathrm{6}}{x}^{\mathrm{3}} \:\leqslant\frac{{sin}\left({n}^{\mathrm{2}} {x}\right)}{{n}}\:\leqslant{nx}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{{N}} \left({nx}−\frac{{n}^{\mathrm{5}} }{\mathrm{6}}\:{x}^{\mathrm{3}} \right)\:\leqslant\:\sum_{{n}=\mathrm{1}} ^{{N}} \:\frac{\left.{sin}\left({n}^{\mathrm{2}} {x}\right)\right)}{{n}}\:\leqslant\sum_{{n}=\mathrm{1}} ^{{N}} {nx}\:\Rightarrow \\ $$$$\frac{{N}\left({N}+\mathrm{1}\right)}{\mathrm{2}}{x}−{x}^{\mathrm{3}} \sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{\mathrm{5}} }{\mathrm{6}}\:\leqslant\:\sum_{{n}=\mathrm{1}} ^{{N}} \:\frac{{sin}\left({n}^{\mathrm{2}} {x}\right)}{{n}}\:\leqslant\frac{{N}\left({N}+\mathrm{1}\right)}{\mathrm{2}}\:{x}\:\:\:\forall{N}\:\in\:{N}\:\Rightarrow \\ $$$$\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\sum_{{n}=\mathrm{1}} ^{{N}} \:\frac{{sin}\left({n}^{\mathrm{2}} {x}\right)}{{n}}\:=\mathrm{0}\:\:{for}\:{all}\:{N}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({n}^{\mathrm{2}} {x}\right)}{{n}}\:=\mathrm{0}\:. \\ $$