# evaluate arithmetic sum by using fourier series

Hi I've been trying for 40 minutes to evaluate the sum of the following arithmetic series with no luck.

$\sum_{n=1}^\infty \frac{sin(2k)}{k}$

I've tried to make this into a fourier series by calculating $c_k$, $b_k$ and $a_k$ but I can't figure out how to proceed. Any advice? :)

$$\sin 2k = \mathfrak{Im}(e^{2ki})$$ so we might hope that $$\sum_{k=1}^{\infty} \frac{\sin 2k}{k} = \mathfrak{Im} \sum_{k=1}^{\infty} \frac{(e^{2i})^k}{k} = - \mathfrak{Im} \ln(1-e^{2i}) = \tan^{-1}\left(\frac{\sin 2}{1-\cos 2}\right)$$
• i used the Maclaurin expansion of $ln(1-x) = -\sum_{k=1}^{\infty} \frac{x^k}{k}$. however i don't know about its convergence on the non-real points of the unit circle – David Holden Oct 3 '14 at 1:19