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Question Number 31984 by abdo imad last updated on 17/Mar/18
$${study}\:{the}\:{covergence}\:{of}\:\:\Sigma\:{u}_{{n}} \:\:{with} \\ $$$${u}_{{n}} =^{{n}} \sqrt{\frac{{n}}{{n}+\mathrm{1}}}\:−\mathrm{1}\:\:\:. \\ $$
Commented by prof Abdo imad last updated on 22/Mar/18
$${we}\:{have}\:\:{u}_{{n}} \:=\:\left(\:\frac{{n}}{{n}\:+\mathrm{1}}\right)^{\frac{\mathrm{1}}{{n}}} \:−\mathrm{1} \\ $$$$=\:\left(\mathrm{1}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)^{\frac{\mathrm{1}}{{n}}} \:−\mathrm{1}\:\:\:{but}\:{we}\:{have}\: \\ $$$$\left(\mathrm{1}+{x}\right)^{\alpha} \:\sim\:\mathrm{1}\:+\:\alpha{x}\:{for}\:{x}\in{V}\left(\mathrm{0}\right)\:\Rightarrow \\ $$$$\left(\mathrm{1}\:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)^{\frac{\mathrm{1}}{{n}}} \:\sim\mathrm{1}−\:\frac{\mathrm{1}}{{n}\left({n}\:+\mathrm{1}\right)}\:\Rightarrow \\ $$$${u}_{{n}} \:\:\sim\:\:\:−\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}\:\sim\:−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\:{due}\:{to}\:{convergence}\:{of} \\ $$$$\Sigma\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\:{the}\:{serie}\:\:\Sigma{u}_{{n}} {converges}. \\ $$