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Question Number 184932 by mathlove last updated on 14/Jan/23

Answered by qaz last updated on 14/Jan/23

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}\sin \frac{n\pi }{3}=\mathrm{\Im }\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{e^{i\frac{n\pi }{3}}}{n}=-\mathrm{\Im }ln(1-e^{\frac{i\pi }{3}})$$$$=-\mathrm{\Im }ln(\frac{1}{2}-\frac{\sqrt{3}}{2}i)=-\mathrm{\Im }{lne}^{-i\arctan \sqrt{3}}=\frac{\pi }{3}$$

Commented by mathlove last updated on 20/Jan/23

$$whatsthe\mathrm{\Im }??$$

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