Question Number 132162 by Dwaipayan Shikari last updated on 11/Feb/21
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({n}\right)}{{n}^{\mathrm{2}} } \\ $$
Answered by mnjuly1970 last updated on 11/Feb/21
![(1/(2i))[li_2 (e^i )−li_2 (e^(−i) )]](https://www.tinkutara.com/question/Q132166.png)
$$\frac{\mathrm{1}}{\mathrm{2}{i}}\left[{li}_{\mathrm{2}} \left({e}^{{i}} \right)−{li}_{\mathrm{2}} \left({e}^{−{i}} \right)\right] \\ $$
Commented by Dwaipayan Shikari last updated on 11/Feb/21
$${But}\:{exact}\:{result}\:{possible}\:{sir}? \\ $$
Commented by mnjuly1970 last updated on 11/Feb/21
$$\:{i}\:{will}\:{try}\:{sir}\:{payan} \\ $$$$\:\Sigma\frac{{cos}\left({n}\right)}{{n}^{\mathrm{2}} }\:{is}\:{very}\:{simple}\:. \\ $$$$\Sigma\frac{{sin}\left({n}\right)}{{n}^{\mathrm{2}} }\:{is}\:{a}\:{little}\:{bit}\:{challanging} \\ $$$$\:{fourier}\:{series}\:{is}\:{useful}.. \\ $$