Question-173403

Question Number 173403 by mathlove last updated on 11/Jul/22

Commented by mr W last updated on 11/Jul/22

(((n^2 )^(1/3) sin (n!))/(n+1))≥−(n^(2/3) /(n+1))>−(n^(2/3) /n)=−(1/( (n)^(1/3) )) (((n^2 )^(1/3) sin (n!))/(n+1))≤(n^(2/3) /(n+1))<(n^(2/3) /n)=(1/( (n)^(1/3) )) −(1/( (n)^(1/3) ))<(((n^2 )^(1/3) sin (n!))/(n+1))<(1/( (n)^(1/3) )) lim_(n→∞) (−(1/( (n)^(1/3) )))<lim_(n→∞) (((n^2 )^(1/3) sin (n!))/(n+1))<lim_(n→∞) (1/( (n)^(1/3) )) 0<lim_(n→∞) (((n^2 )^(1/3) sin (n!))/(n+1))<0 ⇒lim_(n→∞) (((n^2 )^(1/3) sin (n!))/(n+1))=0 ✓

$$\frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }\:\mathrm{sin}\:\left({n}!\right)}{{n}+\mathrm{1}}\geqslant−\frac{{n}^{\frac{\mathrm{2}}{\mathrm{3}}} }{{n}+\mathrm{1}}>−\frac{{n}^{\frac{\mathrm{2}}{\mathrm{3}}} }{{n}}=−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{n}}} \\ $$$$\frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }\:\mathrm{sin}\:\left({n}!\right)}{{n}+\mathrm{1}}\leqslant\frac{{n}^{\frac{\mathrm{2}}{\mathrm{3}}} }{{n}+\mathrm{1}}<\frac{{n}^{\frac{\mathrm{2}}{\mathrm{3}}} }{{n}}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{n}}} \\ $$$$−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{n}}}<\frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }\:\mathrm{sin}\:\left({n}!\right)}{{n}+\mathrm{1}}<\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{n}}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(−\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{n}}}\right)<\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }\:\mathrm{sin}\:\left({n}!\right)}{{n}+\mathrm{1}}<\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{n}}} \\ $$$$\mathrm{0}<\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }\:\mathrm{sin}\:\left({n}!\right)}{{n}+\mathrm{1}}<\mathrm{0} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }\:\mathrm{sin}\:\left({n}!\right)}{{n}+\mathrm{1}}=\mathrm{0}\:\checkmark \\ $$

Commented by mathlove last updated on 11/Jul/22

$$ \\ $$Thank you so much for a long life. I don't have words to thank you for always helping me

Commented by kaivan.ahmadi last updated on 11/Jul/22

−1≤sin(n!)≤1⇒ −((n^2 )^(1/3) /(n+1))≤(((n^2 )^(1/3) sin(n!))/(n+1))≤((n^2 )^(1/3) /(n+1))⇒ lim_(n→∞) −((n^2 )^(1/3) /(n+1))≤lim_(n→∞) (((n^2 )^(1/3) sin(n!))/(n+1))≤lim_(n→∞) ((n^2 )^(1/3) /(n+1)) ⇒0≤lim_(n→∞) (((n^2 )^(1/3) sin(n))/(n+1))≤0 ⇒lim_(n→∞) (((n^2 )^(1/3) sin(n!))/(n+1))=0

$$−\mathrm{1}\leqslant{sin}\left({n}!\right)\leqslant\mathrm{1}\Rightarrow \\ $$$$−\frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }}{{n}+\mathrm{1}}\leqslant\frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }{sin}\left({n}!\right)}{{n}+\mathrm{1}}\leqslant\frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }}{{n}+\mathrm{1}}\Rightarrow \\ $$$${lim}_{{n}\rightarrow\infty} −\frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }}{{n}+\mathrm{1}}\leqslant{lim}_{{n}\rightarrow\infty} \frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }{sin}\left({n}!\right)}{{n}+\mathrm{1}}\leqslant{lim}_{{n}\rightarrow\infty} \frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }}{{n}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{0}\leqslant{lim}_{{n}\rightarrow\infty} \frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }{sin}\left({n}\right)}{{n}+\mathrm{1}}\leqslant\mathrm{0} \\ $$$$\Rightarrow{lim}_{{n}\rightarrow\infty} \frac{\sqrt[{\mathrm{3}}]{{n}^{\mathrm{2}} }{sin}\left({n}!\right)}{{n}+\mathrm{1}}=\mathrm{0} \\ $$

Commented by mathlove last updated on 13/Jul/22

$${thanks} \\ $$

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