What is the closed form of $$\sum^\infty_{n=1}\frac{\sin n}n$$?
My approach:
$$\sum^\infty_{n=1}\frac{\sin n}n=\Im\left[\sum^\infty_{n=1}\frac{e^{in}}n\right]=\Im\left[-\text{Log}(1-e^{i})\right]$$ where $\text{Log}(1):=0$.
$$\Im\left[-\text{Log}(1-e^i)\right]=-\arg(1-e^i)=-\frac{1-\pi}2$$
Thus, $$\sum^\infty_{n=1}\frac{\sin n}n=\frac{\pi-1}2$$
And as a bonus, $$\sum^\infty_{n=1}\frac{\cos n}n=\ln(2\sec1)$$
Is this correct?
