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Question Number 139479 by mnjuly1970 last updated on 27/Apr/21

                             ....advanced ....★★★.....calculus.....              :=Σ_(n=1) ^∞ (((sin(n))/n))^3 =?

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....{advanced}\:....\bigstar\bigstar\bigstar.....{calculus}..... \\ $$$$\:\:\:\:\:\:\:\:\:\:\: :=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{sin}\left({n}\right)}{{n}}\right)^{\mathrm{3}} =?\: \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$

Answered by Dwaipayan Shikari last updated on 27/Apr/21

Σ_(n=1) ^∞ ((sin^3 (n))/n^3 )=(3/4)Σ_(n=1) ^∞ ((sin(n))/n^3 )−(1/4)Σ_(n=1) ^∞ ((sin(3n))/n^3 )  =(3/4)((1/(12))+(π^2 /6)−(π/4))−(1/4)((3/(12))+(π^2 /2)−((9π)/4))  =((9π)/(16))−((3π)/(16))=((3π)/8)      (  Σ_(n=1) ^∞ ((sin(nθ))/n^3 )=(θ^3 /(12))+((π^2 θ)/6)−((πθ^2 )/4))

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}^{\mathrm{3}} \left({n}\right)}{{n}^{\mathrm{3}} }=\frac{\mathrm{3}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({n}\right)}{{n}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left(\mathrm{3}{n}\right)}{{n}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{12}}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\pi}{\mathrm{4}}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{3}}{\mathrm{12}}+\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{9}\pi}{\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{9}\pi}{\mathrm{16}}−\frac{\mathrm{3}\pi}{\mathrm{16}}=\frac{\mathrm{3}\pi}{\mathrm{8}}\:\:\:\:\:\:\left(\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({n}\theta\right)}{{n}^{\mathrm{3}} }=\frac{\theta^{\mathrm{3}} }{\mathrm{12}}+\frac{\pi^{\mathrm{2}} \theta}{\mathrm{6}}−\frac{\pi\theta^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$

Commented by mnjuly1970 last updated on 27/Apr/21

 mercey mr Dwaipayan..

$$\:{mercey}\:{mr}\:{Dwaipayan}.. \\ $$

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