Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 32037 by abdo imad last updated on 18/Mar/18

let u_n =cos(π(√(n^2 +n+1))) find nature of Σ u_n .

$${let}\:\:{u}_{{n}} ={cos}\left(\pi\sqrt{{n}^{\mathrm{2}} \:+{n}+\mathrm{1}}\right)\:{find}\:{nature}\:{of}\:\Sigma\:{u}_{{n}} . \\ $$$$ \\ $$

Commented by abdo imad last updated on 22/Mar/18

we have π(√(n^2 +n+1)) =πn(√(1+(1/n) +(1/n^2 ))) ∼nπ(1+ (1/2)((1/n) +(1/n^2 ))=nπ +(π/2) +(π/n) ⇒ cos(π(√(n^2 +n+1 ))) ∼cos(nπ +(π/2) +(π/n))=−sin(nπ +(π/n)) =(−1)^(n−1) cos((π/n)) but we have cos((π/n))∼1−(π^2 /n^2 ) ⇒ u_n ∼ (−1)^(n−1) (1−(π^2 /n^2 )) but Σ (−1)^n diverges ⇒Σu_(n ) diverges

$${we}\:{have}\:\pi\sqrt{{n}^{\mathrm{2}} \:+{n}+\mathrm{1}}\:=\pi{n}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}}\:+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }} \\ $$$$\sim{n}\pi\left(\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{n}}\:+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)={n}\pi\:+\frac{\pi}{\mathrm{2}}\:\:+\frac{\pi}{{n}}\:\Rightarrow\right. \\ $$$${cos}\left(\pi\sqrt{{n}^{\mathrm{2}} \:+{n}+\mathrm{1}\:}\right)\:\sim{cos}\left({n}\pi\:+\frac{\pi}{\mathrm{2}}\:+\frac{\pi}{{n}}\right)=−{sin}\left({n}\pi\:+\frac{\pi}{{n}}\right) \\ $$$$=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:{cos}\left(\frac{\pi}{{n}}\right)\:{but}\:{we}\:{have}\:{cos}\left(\frac{\pi}{{n}}\right)\sim\mathrm{1}−\frac{\pi^{\mathrm{2}} }{{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$${u}_{{n}} \sim\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left(\mathrm{1}−\frac{\pi^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\:{but}\:\Sigma\:\left(−\mathrm{1}\right)^{{n}} \:{diverges}\:\Rightarrow\Sigma{u}_{{n}\:} \:{diverges} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com