Question Number 32037 by abdo imad last updated on 18/Mar/18
$${let}\:\:{u}_{{n}} ={cos}\left(\pi\sqrt{{n}^{\mathrm{2}} \:+{n}+\mathrm{1}}\right)\:{find}\:{nature}\:{of}\:\Sigma\:{u}_{{n}} . \\ $$$$ \\ $$
Commented by abdo imad last updated on 22/Mar/18
$${we}\:{have}\:\pi\sqrt{{n}^{\mathrm{2}} \:+{n}+\mathrm{1}}\:=\pi{n}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}}\:+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }} \\ $$$$\sim{n}\pi\left(\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{n}}\:+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)={n}\pi\:+\frac{\pi}{\mathrm{2}}\:\:+\frac{\pi}{{n}}\:\Rightarrow\right. \\ $$$${cos}\left(\pi\sqrt{{n}^{\mathrm{2}} \:+{n}+\mathrm{1}\:}\right)\:\sim{cos}\left({n}\pi\:+\frac{\pi}{\mathrm{2}}\:+\frac{\pi}{{n}}\right)=−{sin}\left({n}\pi\:+\frac{\pi}{{n}}\right) \\ $$$$=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{cos}\left(\frac{\pi}{{n}}\right)\:{but}\:{we}\:{have}\:{cos}\left(\frac{\pi}{{n}}\right)\sim\mathrm{1}−\frac{\pi^{\mathrm{2}} }{{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$${u}_{{n}} \sim\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left(\mathrm{1}−\frac{\pi^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\:{but}\:\Sigma\:\left(−\mathrm{1}\right)^{{n}} \:{diverges}\:\Rightarrow\Sigma{u}_{{n}\:} \:{diverges} \\ $$