Question Number 164475 by mathocean1 last updated on 17/Jan/22
$${Etudiez}\:\:{la}\:{convergence}\:{de}\:{de}\:{la} \\ $$$${suite}\:{U}_{{n}} =\frac{\mathrm{1}+{cos}\left({n}\right)+\mathrm{2}{n}}{{ni}+\sqrt{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}}\:;\:{n}\:\in\:\mathbb{N}. \\ $$$$\left[{study}\:{the}\:{convergence}\:{of}\:{U}_{{n}} \right] \\ $$
Answered by puissant last updated on 19/Jan/22
$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{U}_{{n}} \:=\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:\frac{{n}\left(\frac{\mathrm{1}}{{n}}+\frac{{cos}\left({n}\right)}{{n}}+\mathrm{2}\right)}{{n}\left({i}+\sqrt{\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)}\right.} \\ $$$$\:\Rightarrow\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{U}_{{n}} \:=\:\frac{\mathrm{2}}{{i}+\mathrm{1}}\:=\:\frac{\mathrm{2}\left(\mathrm{1}−{i}\right)}{\:\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}}−\sqrt{\mathrm{2}}{i}. \\ $$
Commented by mathocean1 last updated on 19/Jan/22
$${thanks} \\ $$