A expression of sine

Expansion of \ln(1-z)

\sum_{n=1}^{\infty} \frac{z^n}{n}=\ln(1-z)

Hence, \sum_{n=1}^{\infty} \frac{\sin(n\theta)}{n} = \sum_{n=1}^{\infty}\frac{e^{in\theta}-e^{-in\theta}}{2in} = \frac{\ln(1-e^{i\theta})-\ln(1-e^{-i\theta})}{2i} = \frac{\pi-\theta}{2}.

We can apply above formula: \sum_{n=1}^{\infty} \frac{\sin^3 n}{n} = \sum_{n=1}^{\infty} \frac{3\sin n -\sin(3n)}{4n} = \frac34\cdot\frac{\pi-1}{2} - \frac14\cdot\frac{\pi-3}{2} = \frac{\pi}{4}.

https://artofproblemsolving.com/community/c7t443f7h1390328_convergence_of_sine_series

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