Question Number 116166 by Ar Brandon last updated on 01/Oct/20
$$\forall{n}\in\mathbb{N}^{\ast} ,\:\mathrm{suppose}\:{u}_{{n}} =\left(\mathrm{5sin}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{5}}\mathrm{cos}\:{n}\right)^{{n}} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}{u}_{{n}} =\mathrm{0} \\ $$
Answered by mindispower last updated on 01/Oct/20
$${since}\:{sin}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\rightarrow\mathrm{0}\Rightarrow\exists{N}\:{such}\:\forall{n}\geqslant{N} \\ $$$$\mathrm{0}\leqslant{sin}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)<\frac{\mathrm{1}}{\mathrm{10}} \\ $$$$\forall{x}\in\mathbb{R}\:\:\:\:−\mathrm{1}\leqslant{cos}\left({x}\right)\leqslant\mathrm{1} \\ $$$$\Rightarrow\forall{n}\geqslant{N} \\ $$$$\mathrm{0}−\frac{\mathrm{1}}{\mathrm{5}}\leqslant\mathrm{5}{sin}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)+\frac{{cos}\left({n}\right)}{\mathrm{5}}\leqslant\frac{\mathrm{5}}{\mathrm{10}}+\frac{\mathrm{1}}{\mathrm{5}}=\frac{\mathrm{7}}{\mathrm{10}} \\ $$$$\forall{n}\geqslant{N} \\ $$$$\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{5}}\leqslant\mathrm{5}{sin}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)+\frac{{cos}\left({n}\right)}{\mathrm{5}}\leqslant\frac{\mathrm{7}}{\mathrm{10}} \\ $$$$\Rightarrow−\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{{n}} \leqslant\left(\mathrm{5}{sin}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)+\frac{{cos}\left({n}\right)}{\mathrm{5}}\right)^{{n}} \leqslant\left(\frac{\mathrm{7}}{\mathrm{10}}\right)^{{n}} \\ $$$${since}\:−\left(\frac{\mathrm{1}}{\mathrm{5}}\right)^{{n}} \rightarrow\mathrm{0},\left(\frac{\mathrm{7}}{\mathrm{10}}\right)^{{n}} \rightarrow\mathrm{0} \\ $$$${we}\:{get}\:{our}\:{Ansewr} \\ $$$$ \\ $$$$ \\ $$