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

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

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

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

Commented by mnjuly1970 last updated on 27/Apr/21

$$merceymrDwaipayan..$$

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