Question Number 40149 by maxmathsup by imad last updated on 16/Jul/18
$${let}\:{u}_{{n}} =\:\frac{\mathrm{1}}{\sqrt{{n}}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\sqrt{{n}+\mathrm{4}{k}}} \\ $$$${find}\:{lim}_{{n}\rightarrow+\infty} \:{u}_{{n}} \\ $$
Commented by math khazana by abdo last updated on 17/Jul/18
![we have u_n = (1/n) Σ_(k=1) ^n (1/(√(1+4(k/n)))) , u_n is a Rieman sum and lim_(n→+∞) u_n = ∫_0 ^1 (dx/(√(1+4x))) =[(1/2)(√(1+4x))]_0 ^1 =(1/2)((√5) −1)](Q40204.png)
$${we}\:{have}\:{u}_{{n}} =\:\frac{\mathrm{1}}{{n}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{4}\frac{{k}}{{n}}}}\:\:,\:{u}_{{n}} \:{is}\:{a}\:{Rieman} \\ $$$${sum}\:\:{and}\:{lim}_{{n}\rightarrow+\infty} \:{u}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{\sqrt{\mathrm{1}+\mathrm{4}{x}}} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+\mathrm{4}{x}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{5}}\:\:−\mathrm{1}\right)\: \\ $$$$ \\ $$