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Arithmetic Questions

Question Number 100297 by Coronavirus last updated on 26/Jun/20

$${calulate}\:{using}\:{Riemann}\:{sums} \\$$$${tbe}\:{limit}\:{of}\:{this}\:{sequence} \\$$$$\:\:\underset{{k}={n}} {\overset{\mathrm{2}{n}} {\sum}}\mathrm{sin}\:\left(\frac{\pi}{{k}}\right) \\$$

Answered by abdomsup last updated on 26/Jun/20

$${A}_{{n}} =_{{p}={k}−{n}} \:\:\sum_{{p}=\mathrm{0}} ^{{n}} \:{sin}\left(\frac{\pi}{{p}+{n}}\right) \\$$$${we}\:{have}\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\leqslant{sinx}\leqslant{x}\:\Rightarrow \\$$$$\frac{\pi}{{k}}−\frac{\pi^{\mathrm{3}} }{\mathrm{6}{k}^{\mathrm{3}} }\leqslant{sin}\left(\frac{\pi}{{k}}\right)\leqslant\frac{\pi}{{k}} \\$$$$\Rightarrow\sum_{{k}={n}} ^{\mathrm{2}{n}} \:\frac{\pi}{{k}}−\pi^{\mathrm{3}} \:\sum_{{k}={n}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{3}} } \\$$$$\leqslant\sum_{{k}={n}} ^{\mathrm{2}{n}} \:{sin}\left(\frac{\pi}{{k}}\right)\leqslant\pi\sum_{{k}={n}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}} \\$$$$\pi\:\sum_{{p}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{{p}+{n}}−\pi^{\mathrm{3}\:} \sum_{{p}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\left({p}+{n}\right)^{\mathrm{3}} } \\$$$$\leqslant\sum_{{k}={n}} ^{\mathrm{2}{n}} \:{sin}\left(\frac{\pi}{{k}}\right)\leqslant\pi\sum_{{p}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{{p}+{n}} \\$$$${we}\:{have}\sum_{{p}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{{p}+{n}} \\$$$$\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}+....+\frac{\mathrm{1}}{\mathrm{2}{n}} \\$$$$\mathrm{1}+\frac{\mathrm{1}}{{n}}+....+\frac{\mathrm{1}}{{n}−\mathrm{1}}+....+\frac{\mathrm{1}}{\mathrm{2}{n}} \\$$$$−{H}_{{n}−\mathrm{1}} ={H}_{\mathrm{2}{n}} −{H}_{{n}−\mathrm{1}} \\$$$$={ln}\left(\mathrm{2}{n}\right)+\gamma\:+\mathrm{0}\left(\frac{\mathrm{1}}{\mathrm{2}{n}}\right)−{ln}\left({n}−\mathrm{1}\right) \\$$$$−\gamma−{o}\left(\frac{\mathrm{1}}{{n}−\mathrm{1}}\right)\rightarrow{ln}\mathrm{2} \\$$$${its}\:{clear}\:\:{that}\:\sum_{{p}=\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{\left({p}+{n}\right)^{\mathrm{3}} }\rightarrow\mathrm{0} \\$$$$\left.{b}\left.{ecause}\:\Sigma\right)...\right)\leqslant\sum_{\mathrm{0}} ^{{n}} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }=\frac{{n}+\mathrm{1}}{{n}^{\mathrm{3}} } \\$$$$\Rightarrow{lim}_{{n}\rightarrow+\infty} \sum_{{k}={n}} ^{\mathrm{2}{n}} \:{sin}\left(\frac{{k}}{{n}}\right)=\pi{ln}\mathrm{2} \\$$

Commented by Coronavirus last updated on 26/Jun/20

Thanks you Mr

Commented by DGmichael last updated on 26/Jun/20

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Commented by mathmax by abdo last updated on 26/Jun/20

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir} \\$$

Commented by Ar Brandon last updated on 26/Jun/20

Qui a compris ? ����

Commented by Ar Brandon last updated on 26/Jun/20

$$\mathrm{Sir}\:\mathrm{is}\:\mathrm{H}_{\mathrm{n}} \:\mathrm{a}\:\mathrm{special}\:\mathrm{function}\:? \\$$

Commented by mathmax by abdo last updated on 26/Jun/20

$$\mathrm{H}_{\mathrm{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\:+...+\frac{\mathrm{1}}{\mathrm{n}}\:\mathrm{is}\:\mathrm{harmonic}\:\mathrm{sequence} \\$$

Commented by Ar Brandon last updated on 27/Jun/20

OK thank you Sir

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