Question Number 184932 by mathlove last updated on 14/Jan/23
Answered by qaz last updated on 14/Jan/23
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\mathrm{sin}\:\frac{{n}\pi}{\mathrm{3}}=\Im\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{e}^{{i}\frac{{n}\pi}{\mathrm{3}}} }{{n}}=−\Im{ln}\left(\mathrm{1}−{e}^{\frac{{i}\pi}{\mathrm{3}}} \right) \\ $$$$=−\Im{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\right)=−\Im{lne}^{−{i}\mathrm{arctan}\:\sqrt{\mathrm{3}}} =\frac{\pi}{\mathrm{3}} \\ $$
Commented by mathlove last updated on 20/Jan/23
$${whats}\:{the}\:\Im?? \\ $$