Previous in Relation and Functions Next in Relation and Functions
Question Number 31521 by abdo imad last updated on 09/Mar/18
$${study}\:{the}\:{convergence}\:{of}\:{u}_{{n}} =\sqrt{{n}+\mathrm{1}}\:−\sqrt{{n}−\mathrm{1}}\: \\ $$
Commented by abdo imad last updated on 12/Mar/18
$${we}\:{have}\:{u}_{{n}} =\sqrt{{n}}\left(\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}}}\:−\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{n}}}\:\right)\:{but} \\ $$$$\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{n}}}\:\sim\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}{n}}\:\:{and}\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{n}}}\:\:\sim\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}}\:\Rightarrow \\ $$$${u}_{{n}} \sim\sqrt{{n}}\:\left(\:\frac{\mathrm{1}}{{n}}\right)\:\:\Rightarrow\:{u}_{{n}} \sim\:\:\frac{\mathrm{1}}{\sqrt{{n}}}\:\left({n}\rightarrow\infty\right)\:{so}\:\left({u}_{{n}} \right)\:{is}\:{convergent} \\ $$$${and}\:{lim}_{{n}\rightarrow\infty} {u}_{{n}} \:\:=\mathrm{0}. \\ $$