This question already has an answer here:

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 $\text{Log}(1):=0$.


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?


marked as duplicate by Winther, Gerry Myerson, Adrian Keister, Delta-u, Strants Sep 10 '18 at 17:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.