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

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