Question Number 32041 by abdo imad last updated on 18/Mar/18
$${find}\:{the}\:{nature}\:{of}\:\Sigma\:{u}_{{n}} \:\:/ \\ $$$${u}_{{n}} =\:\frac{\sqrt{\mathrm{1}}\:+\sqrt{\mathrm{2}}\:+….+\sqrt{{n}}}{{n}^{\mathrm{3}} }\:. \\ $$
Commented by abdo imad last updated on 20/Mar/18
![we have u_n = (1/n^2 ) ( (1/( (√n))) Σ_(k=1) ^n (√( (k/n))) ) =(1/(n(√n)))( (1/n) Σ_(k=1) ^n (√(k/n)) ) but lim_(n→∞) (1/n) Σ_(k=1) ^n (√(k/n)) = ∫_0 ^1 (√x) dx = [ (2/3) x^(3/2) ]_0 ^1 = (2/3) ⇒ u_n ∼ (2/(3n(√n))) and the serie Σ_(n≥1) (2/(3n(√n))) is convergent so Σ u_(n ) converges.](https://www.tinkutara.com/question/Q32122.png)
$${we}\:{have}\:{u}_{{n}} =\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\left(\:\frac{\mathrm{1}}{\:\sqrt{{n}}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\sqrt{\:\frac{{k}}{{n}}}\:\right) \\ $$$$=\frac{\mathrm{1}}{{n}\sqrt{{n}}}\left(\:\frac{\mathrm{1}}{{n}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\sqrt{\frac{{k}}{{n}}}\:\right)\:{but}\:{lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\sqrt{\frac{{k}}{{n}}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{{x}}\:{dx}\:=\:\left[\:\frac{\mathrm{2}}{\mathrm{3}}\:{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \:\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\:\Rightarrow\:\:{u}_{{n}} \:\sim\:\:\frac{\mathrm{2}}{\mathrm{3}{n}\sqrt{{n}}}\:\:{and} \\ $$$${the}\:{serie}\:\:\sum_{{n}\geqslant\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}}{\mathrm{3}{n}\sqrt{{n}}}\:{is}\:{convergent}\:{so}\:\Sigma\:{u}_{{n}\:} {converges}. \\ $$