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Question Number 133432 by Dwaipayan Shikari last updated on 22/Feb/21

((sin(√π))/1^3 )+((sin(√(4π)))/2^3 )+((sin(√(9π)))/3^3 )+((sin(√(16π)))/4^3 )+....=((π(√π))/(12))(1−3(√π)+2π)

$$\frac{{sin}\sqrt{\pi}}{\mathrm{1}^{\mathrm{3}} }+\frac{{sin}\sqrt{\mathrm{4}\pi}}{\mathrm{2}^{\mathrm{3}} }+\frac{{sin}\sqrt{\mathrm{9}\pi}}{\mathrm{3}^{\mathrm{3}} }+\frac{{sin}\sqrt{\mathrm{16}\pi}}{\mathrm{4}^{\mathrm{3}} }+....=\frac{\pi\sqrt{\pi}}{\mathrm{12}}\left(\mathrm{1}−\mathrm{3}\sqrt{\pi}+\mathrm{2}\pi\right) \\ $$

Answered by mindispower last updated on 24/Feb/21

Σ_(n≥0) ((sim(nx))/n)=((π−x)/2_ ) ⇒Σ((cos(nx))/n^2 )=−(π/2)x+(x^2 /4)+(π^2 /6) ⇒Σ_(n≥1) ((sin(nx))/n^3 )=−((πx^2 )/4)+(x^3 /(12))+(π^2 /6)x=f(x) ((sin((√π)))/1^3 )+((sin((√(4π))))/2^2 )+.....+((sin((√(n^2 π))))/n^2 )+... =Σ((sin(n(√π)))/n^3 )=f((√π))=−(π^2 /4)+((π(√π))/(12))+((π^2 (√π))/6) =((π(√π))/(12))(1−3(√π)+2π)

$$\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{{sim}\left({nx}\right)}{{n}}=\frac{\pi−{x}}{\mathrm{2}_{} } \\ $$$$\Rightarrow\Sigma\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} }=−\frac{\pi}{\mathrm{2}}{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Rightarrow\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{sin}\left({nx}\right)}{{n}^{\mathrm{3}} }=−\frac{\pi{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{{x}^{\mathrm{3}} }{\mathrm{12}}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}{x}={f}\left({x}\right) \\ $$$$\frac{{sin}\left(\sqrt{\pi}\right)}{\mathrm{1}^{\mathrm{3}} }+\frac{{sin}\left(\sqrt{\mathrm{4}\pi}\right)}{\mathrm{2}^{\mathrm{2}} }+.....+\frac{{sin}\left(\sqrt{{n}^{\mathrm{2}} \pi}\right)}{{n}^{\mathrm{2}} }+... \\ $$$$=\Sigma\frac{{sin}\left({n}\sqrt{\pi}\right)}{{n}^{\mathrm{3}} }={f}\left(\sqrt{\pi}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}+\frac{\pi\sqrt{\pi}}{\mathrm{12}}+\frac{\pi^{\mathrm{2}} \sqrt{\pi}}{\mathrm{6}} \\ $$$$=\frac{\pi\sqrt{\pi}}{\mathrm{12}}\left(\mathrm{1}−\mathrm{3}\sqrt{\pi}+\mathrm{2}\pi\right) \\ $$

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