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

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