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Question Number 32486 by abdo imad last updated on 25/Mar/18

find lim_(n→∞) Σ_(k=n+1) ^(2n) sin((1/k)).

$${find}\:{lim}_{{n}\rightarrow\infty} \:\:\:\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:{sin}\left(\frac{\mathrm{1}}{{k}}\right). \\ $$

Commented by abdo imad last updated on 26/Mar/18

let put S_n = Σ_(k=n+1) ^(2n) sin((1/k))we know that x−(x^3 /6) ≤sinx≤ x ⇒ (1/k) − (1/(6k^3 )) ≤ sin((1/k))≤ (1/k) ⇒ Σ_(k=n+1) ^(2n) (1/k) −(1/6)Σ_(k=n+1) ^(2n) (1/k^3 )≤S_n ≤ Σ_(k=n+1) ^(2n) (1/k) but Σ_(k=n+1) ^(2n) (1/k) =(1/(n+1)) +(1/(n+2)) +.....(1/(2n)) =H_(2n) −H_n H_(2n) =ln(2n) +γ +o(1) H_n =ln(n) +γ +o(1) ⇒ H_(2n) −H_n =ln(((2n)/n)) + o(1)⇒ lim_(n→∞) H_(2n) −H_n =ln(2) also we have n+1≤k≤2n ⇒ (1/(2n))≤ (1/k) ≤ (1/(n+1)) ⇒ (1/(8n^3 )) ≤ (1/k^3 ) ≤ (1/n^3 ) ⇒ (n/(8n^3 )) ≤ Σ_(k=n+1) ^(2n) (1/k^3 )≤ (n/n^3 ) ⇒ (1/(8n^2 )) ≤Σ_(k=n+1) ^(2n) (1/k^3 ) ≤(1/n^2 ) ⇒ lim_(n→∞) Σ_(k=n+1) ^(2n) (1/k^3 ) =0 so lim_(n→∞) S_n =ln(2) .

$${let}\:{put}\:{S}_{{n}} =\:\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} {sin}\left(\frac{\mathrm{1}}{{k}}\right){we}\:{know}\:{that}\:\:\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{6}}\:\leqslant{sinx}\leqslant\:{x}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{k}}\:−\:\frac{\mathrm{1}}{\mathrm{6}{k}^{\mathrm{3}} }\:\leqslant\:{sin}\left(\frac{\mathrm{1}}{{k}}\right)\leqslant\:\frac{\mathrm{1}}{{k}}\:\Rightarrow\:\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}}\:−\frac{\mathrm{1}}{\mathrm{6}}\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\leqslant{S}_{{n}} \leqslant\:\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \frac{\mathrm{1}}{{k}} \\ $$$${but}\:\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}}\:=\frac{\mathrm{1}}{{n}+\mathrm{1}}\:+\frac{\mathrm{1}}{{n}+\mathrm{2}}\:+.....\frac{\mathrm{1}}{\mathrm{2}{n}}\:={H}_{\mathrm{2}{n}} \:−{H}_{{n}} \\ $$$${H}_{\mathrm{2}{n}} ={ln}\left(\mathrm{2}{n}\right)\:+\gamma\:+{o}\left(\mathrm{1}\right) \\ $$$${H}_{{n}} ={ln}\left({n}\right)\:+\gamma\:+{o}\left(\mathrm{1}\right)\:\Rightarrow\:{H}_{\mathrm{2}{n}} −{H}_{{n}} ={ln}\left(\frac{\mathrm{2}{n}}{{n}}\right)\:+\:{o}\left(\mathrm{1}\right)\Rightarrow \\ $$$${lim}_{{n}\rightarrow\infty} {H}_{\mathrm{2}{n}} \:−{H}_{{n}} ={ln}\left(\mathrm{2}\right)\:{also}\:{we}\:{have}\:\:{n}+\mathrm{1}\leqslant{k}\leqslant\mathrm{2}{n}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{n}}\leqslant\:\frac{\mathrm{1}}{{k}}\:\leqslant\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{8}{n}^{\mathrm{3}} }\:\leqslant\:\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\:\leqslant\:\:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:\Rightarrow \\ $$$$\:\frac{{n}}{\mathrm{8}{n}^{\mathrm{3}} }\:\leqslant\:\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\leqslant\:\frac{{n}}{{n}^{\mathrm{3}} }\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{8}{n}^{\mathrm{2}} }\:\leqslant\sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\:\leqslant\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow\infty} \sum_{{k}={n}+\mathrm{1}} ^{\mathrm{2}{n}} \:\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\:=\mathrm{0}\:{so}\:\:{lim}_{{n}\rightarrow\infty} \:{S}_{{n}} \:={ln}\left(\mathrm{2}\right)\:. \\ $$$$ \\ $$

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