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Question Number 97985 by abdomathmax last updated on 10/Jun/20
$$\mathrm{let}\:\mathrm{S}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\sqrt{\mathrm{n}^{\mathrm{2}} +\mathrm{2kn}}} \\ $$$$\mathrm{find}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{S}_{\mathrm{n}} \\ $$
Answered by maths mind last updated on 11/Jun/20
![S_n =(1/n)Σ_(1≤k≤n) (1/(√(1+((2k)/n)))) lim_(n→∞) S_n =∫_0 ^1 (1/(√(1+2x)))−1 =[2(√(1+2x))]_0 ^1 −1 =2(√3)−3](Q98122.png)
$${S}_{{n}} =\frac{\mathrm{1}}{{n}}\underset{\mathrm{1}\leqslant{k}\leqslant{n}} {\sum}\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\frac{\mathrm{2}{k}}{{n}}}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{S}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{2}{x}}}−\mathrm{1} \\ $$$$=\left[\mathrm{2}\sqrt{\mathrm{1}+\mathrm{2}{x}}\right]_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{1} \\ $$$$=\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3} \\ $$
Answered by mathmax by abdo last updated on 12/Jun/20
![S_n =(1/n) Σ_(k=1) ^n (1/(√(1+2(k/n)))) so S_n is a Rieman sum and lim_(n→+∞) S_n =∫_0 ^1 (dx/(√(1+2x))) =[(√(1+2x))]_0 ^1 =(√3)−1](Q98169.png)
$$\mathrm{S}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}}\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{2}\frac{\mathrm{k}}{\mathrm{n}}}}\:\:\mathrm{so}\:\mathrm{S}_{\mathrm{n}} \mathrm{is}\:\mathrm{a}\:\mathrm{Rieman}\:\mathrm{sum}\:\mathrm{and} \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\mathrm{S}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\sqrt{\mathrm{1}+\mathrm{2x}}}\:=\left[\sqrt{\mathrm{1}+\mathrm{2x}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\sqrt{\mathrm{3}}−\mathrm{1} \\ $$