calulate-using-Riemann-sums-tbe-limit-of-this-sequence-k-n-2n-sin-pi-k-

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

calulate using Riemann sums tbe limit of this sequence Σ_(k=n) ^(2n) sin ((π/k))

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

Answered by abdomsup last updated on 26/Jun/20

A_n =_(p=k−n) Σ_(p=0) ^n sin((π/(p+n))) we have x−(x^3 /6)≤sinx≤x ⇒ (π/k)−(π^3 /(6k^3 ))≤sin((π/k))≤(π/k) ⇒Σ_(k=n) ^(2n) (π/k)−π^3 Σ_(k=n) ^(2n) (1/k^3 ) ≤Σ_(k=n) ^(2n) sin((π/k))≤πΣ_(k=n) ^(2n) (1/k) π Σ_(p=0) ^n (1/(p+n))−π^(3 ) Σ_(p=0) ^n (1/((p+n)^3 )) ≤Σ_(k=n) ^(2n) sin((π/k))≤πΣ_(p=0) ^n (1/(p+n)) we haveΣ_(p=0) ^n (1/(p+n)) (1/n)+(1/(n+1))+....+(1/(2n)) 1+(1/n)+....+(1/(n−1))+....+(1/(2n)) −H_(n−1) =H_(2n) −H_(n−1) =ln(2n)+γ +0((1/(2n)))−ln(n−1) −γ−o((1/(n−1)))→ln2 its clear that Σ_(p=0) ^n (1/((p+n)^3 ))→0 because Σ)...)≤Σ_0 ^n (1/n^3 )=((n+1)/n^3 ) ⇒lim_(n→+∞) Σ_(k=n) ^(2n) sin((k/n))=πln2

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

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

H_n =1+(1/2)+(1/3) +...+(1/n) is harmonic sequence

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

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

OK thank you Sir

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