# Closed form of $\sum^\infty_{n=1} \frac{\sin n}n$ [duplicate]

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?

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

• The former sum is correct, but for the latter I find $-\frac{1}{2}\log(2 - 2\cos(1)) = -\log(2 \sin(1/2))$ which differs from what you got (maybe a typo?). See also math.stackexchange.com/a/1612163/147873 – Winther Sep 10 '18 at 12:26