Question Number 31522 by abdo imad last updated on 09/Mar/18
$${let}\:{give}\:{u}_{{n}} =\sqrt{{ln}\left({n}+\mathrm{1}\right)−{ln}\left({n}\right)}\: \\ $$$$\left.\mathrm{1}\right){give}\:{a}\:{simple}\:{eqivalent}\:{of}\:{u}_{{n}} \:\left({n}\rightarrow\infty\right) \\ $$$$\left.\mathrm{2}\right)\:{deduce}\:{the}\:{nature}\:{of}\:{u}_{{n}} . \\ $$
Commented by abdo imad last updated on 15/Mar/18
$${we}\:{have}\:{u}_{{n}} =\left({ln}\left(\frac{{n}+\mathrm{1}}{{n}}\right)\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\left({ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:{but} \\ $$$${ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\:\sim\:\frac{\mathrm{1}}{{n}}\:\left({n}\rightarrow\infty\right)\:\:\Rightarrow\:{u}_{{n}} \:\sim\:\frac{\mathrm{1}}{\sqrt{{n}}}\:. \\ $$$$\left.\mathrm{2}\left.\right)\:{from}\:{Q}.\mathrm{1}\right)\:\left({u}_{{n}} \right)\:{is}\:{convergent}\:{and}\:{lim}_{{n}\rightarrow\infty} {u}_{{n}} =\mathrm{0}\:\:. \\ $$