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Question Number 191369 by mnjuly1970 last updated on 23/Apr/23

$$$$$$calculate$$$$$$$$\mathit{\varphi }=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{sin\left(\frac{n\pi }{3}\right)}{{(2n+1)}^2}=?$$$$$$

Answered by namphamduc last updated on 24/Apr/23

$$S=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{\sin \left(\frac{n\pi }{3}\right)}{{(2n+1)}^2}=\mathrm{\Im }\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\left(e^{\frac{i\pi }{3}}\right)}^n}{{(2n+1)}^2}$$$${\tanh }^{-1}\left(x\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{2n+1}}{2n+1}\Rightarrow \frac{{\tanh }^{-1}\left(x\right)}{x}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{2n}}{2n+1}\Rightarrow \frac{{\tanh }^{-1}\left(x\right)}{x}-1=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{2n}}{2n+1}$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{x^{2n}}{{(2n+1)}^2}=\frac{1}{2x}({\mathrm{Li}}_2\left(x\right)-{\mathrm{Li}}_2(-x))-1$$$$\Rightarrow S=\mathrm{\Im }\left(\frac{1}{2e^{\frac{i\pi }{6}}}({\mathrm{Li}}_2\left(e^{\frac{i\pi }{6}}\right)-{\mathrm{Li}}_2(-e^{\frac{i\pi }{6}}))\right)$$$$=\mathrm{\Im }\left(\frac{1}{(\sqrt{3}+i)}(\frac{{\pi }^2}{6}+i\frac{4}{3}G)\right)=\mathrm{\Im }\left(\frac{1}{4}(\sqrt{3}-i)(\frac{{\pi }^2}{6}+i\frac{4}{3}G)\right)$$$$=-\frac{{\pi }^2}{24}+\frac{G}{\sqrt{3}}$$

Commented by mnjuly1970 last updated on 24/Apr/23

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