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Question Number 121928 by Dwaipayan Shikari last updated on 12/Nov/20

Σ_(n=0) ^∞ ((1/(12n+1))+(1/(12n+5))−(1/(12n+7))−(1/(12n+11)))    Problem source : Brilliant.Org

$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{12}{n}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{12}{n}+\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{12}{n}+\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{12}{n}+\mathrm{11}}\right) \\ $$$$ \\ $$$${Problem}\:{source}\::\:{Brilliant}.{Org} \\ $$

Commented by Dwaipayan Shikari last updated on 12/Nov/20

https://brilliant.org/problems/not-quite-a-telescoping-sum-2 https://brilliant.org/problems/integral-out-of-the-box

Commented by Dwaipayan Shikari last updated on 12/Nov/20

(1/(12))Σ_(n=0) ^∞ ((1/(n+(1/(12))))−(1/(n+((11)/(12))))+(1/(n+(5/(12))))−(1/(n+(7/(12)))))  =(1/(12))Σ_(n=0) ^∞ ((1/(n+(1/(12))))−(1/(n+1−(1/(12)))))+(1/(12))Σ_(n=0) ^∞ ((1/(n+(5/(12))))−(1/(n+1−(7/(12)))))  =(1/(12))(πcot((π/(12)))+πcot(((5π)/(12))))  =(π/(12))(2−(√3)+2+(√3))  =(π/3)

$$\frac{\mathrm{1}}{\mathrm{12}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}+\frac{\mathrm{1}}{\mathrm{12}}}−\frac{\mathrm{1}}{{n}+\frac{\mathrm{11}}{\mathrm{12}}}+\frac{\mathrm{1}}{{n}+\frac{\mathrm{5}}{\mathrm{12}}}−\frac{\mathrm{1}}{{n}+\frac{\mathrm{7}}{\mathrm{12}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}+\frac{\mathrm{1}}{\mathrm{12}}}−\frac{\mathrm{1}}{{n}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{12}}}\right)+\frac{\mathrm{1}}{\mathrm{12}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}+\frac{\mathrm{5}}{\mathrm{12}}}−\frac{\mathrm{1}}{{n}+\mathrm{1}−\frac{\mathrm{7}}{\mathrm{12}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left(\pi{cot}\left(\frac{\pi}{\mathrm{12}}\right)+\pi{cot}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)\right) \\ $$$$=\frac{\pi}{\mathrm{12}}\left(\mathrm{2}−\sqrt{\mathrm{3}}+\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$$=\frac{\pi}{\mathrm{3}} \\ $$

Answered by mnjuly1970 last updated on 12/Nov/20

solution:  S=Σ_(n≥0) ((1/(12n+1))−(1/(12n+11)))+Σ_(n≥0) ((1/(12n+5))−(1/(12n+7)))    =(1/(12))Σ_(n≥1) ((1/(n−((11)/(12))))−(1/(n−(1/(12)))))+(1/(12))Σ_(n≥1) ((1/(n−(7/(12))))−(1/(n−(5/(12)))))  =(1/(12))(ψ(((11)/(12)))−ψ((1/(12))))+(1/(12))(ψ((7/(12 )))−ψ((5/(12))))   we know that:     ψ(1−s)−ψ(s)=πcot(πs)  ∴  S=(π/(12))(cot((π/(12)))+cot(((5π)/(12))))        note:: tan(15^° )=((1−((√3)/3))/(1+((√3)/3)))=((12−6(√3))/6)=2−(√3)    S=(π/(12))(2+(√3) +2−(√3) )=(π/3)      i don′t know finall answer     is correct  or no  please tell me.

$${solution}: \\ $$$${S}=\underset{{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{12}{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{12}{n}+\mathrm{11}}\right)+\underset{{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{\mathrm{12}{n}+\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{12}{n}+\mathrm{7}}\right) \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{12}}\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{{n}−\frac{\mathrm{11}}{\mathrm{12}}}−\frac{\mathrm{1}}{{n}−\frac{\mathrm{1}}{\mathrm{12}}}\right)+\frac{\mathrm{1}}{\mathrm{12}}\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{{n}−\frac{\mathrm{7}}{\mathrm{12}}}−\frac{\mathrm{1}}{{n}−\frac{\mathrm{5}}{\mathrm{12}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left(\psi\left(\frac{\mathrm{11}}{\mathrm{12}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{12}}\right)\right)+\frac{\mathrm{1}}{\mathrm{12}}\left(\psi\left(\frac{\mathrm{7}}{\mathrm{12}\:}\right)−\psi\left(\frac{\mathrm{5}}{\mathrm{12}}\right)\right) \\ $$$$\:{we}\:{know}\:{that}: \\ $$$$\:\:\:\psi\left(\mathrm{1}−{s}\right)−\psi\left({s}\right)=\pi{cot}\left(\pi{s}\right) \\ $$$$\therefore\:\:{S}=\frac{\pi}{\mathrm{12}}\left({cot}\left(\frac{\pi}{\mathrm{12}}\right)+{cot}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)\right) \\ $$$$\:\:\:\:\:\:{note}::\:{tan}\left(\mathrm{15}^{°} \right)=\frac{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}{\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}=\frac{\mathrm{12}−\mathrm{6}\sqrt{\mathrm{3}}}{\mathrm{6}}=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\:\:{S}=\frac{\pi}{\mathrm{12}}\left(\mathrm{2}+\sqrt{\mathrm{3}}\:+\mathrm{2}−\sqrt{\mathrm{3}}\:\right)=\frac{\pi}{\mathrm{3}} \\ $$$$\:\:\:\:{i}\:{don}'{t}\:{know}\:{finall}\:{answer} \\ $$$$\:\:\:{is}\:{correct}\:\:{or}\:{no}\:\:{please}\:{tell}\:{me}. \\ $$$$ \\ $$

Commented by Dwaipayan Shikari last updated on 12/Nov/20

Yes it is correct sir!

$${Yes}\:{it}\:{is}\:{correct}\:{sir}! \\ $$

Commented by mnjuly1970 last updated on 12/Nov/20

thank you so much  your questions is very nice   grateful ....

$${thank}\:{you}\:{so}\:{much} \\ $$$${your}\:{questions}\:{is}\:{very}\:{nice}\: \\ $$$${grateful}\:.... \\ $$

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