I recently learnt that sums of sine and cosine series whose arguments are in arithmetic progression can be evaluated using complex numbers. I modified it a little; like below I divided it by the argument of sine, and didn't expected a closed form sum. But surprisingly , Wolfram calculator gives this nice answer for it's sum. $$\sum_{n≥1}\frac{\sin n}{n}= \frac{1}{2}(\pi-1)$$
I've never encountered such series so I'm interested in knowing that what are the methods using which we can prove these sums ( assuming the result given by calculator as true ). Thanks !
We have that $$\sum_{n=1}^\infty\frac{z^n}n=-\ln(1-z)$$ for $|z|<1.$
I think due to Dirichlet test, we can use this formula for $z=e^{ix}\neq1$. Then,
$$\begin{align} \sum_{n=1}^\infty\frac{\sin nx}n &= \frac1{2i}\left(\sum_{n=1}^\infty\frac{e^{ixn}}n-\sum_{n=1}^\infty\frac{e^{-ixn}}n\right)\\\\ &=\frac1{2i}\ln\left(\frac{1-e^{-ix}}{1-e^{ix}}\right)\\\\ &=\frac{\pi-x}{2} \end{align}$$ which is true for $0<x<2\pi$.