Difficult infinite trigonometric series Evaluate the sum of the following infinite series. 
$$\left(\sin{\frac{\pi}{3}}\right) + \left(\frac{1}{2}\sin{\frac{2\pi}{3}}\right) + \left(\frac{1}{3}\sin{\frac{3\pi}{3}}\right) + \ldots$$

The result is none other than $\dfrac\pi3$ , with the obvious generalization $\displaystyle\sum_{n=1}^\infty\frac{\sin(nx)}n=\frac{\pi-x^*}2$, where $x^*=x\mod2\pi.~$ Unfortunately, the only way in which I am able to “prove” this is by ascribing finite values to divergent series by way of analytic continuation, which isn't really “kosher”, so $...$
