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? :)
just a thought...
reasoning formally (i.e. leaving aside delicate questions of convergence)
$$ \sin 2k = \mathfrak{Im}(e^{2ki}) $$ so we might hope that $$ \sum_{k=1}^{\infty} \frac{\sin 2k}{k} = \mathfrak{Im} \sum_{k=1}^{\infty} \frac{(e^{2i})^k}{k} = - \mathfrak{Im} \ln(1-e^{2i}) = \tan^{-1}\left(\frac{\sin 2}{1-\cos 2}\right) $$