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 $$ \sum_{k=1}^{\infty} \frac{\sin 2k}{k} = \mathfrak{Im} \sum_{k=1}^{\infty} \frac{(e^{2i})^k}{k} = - \mathfrak{Im} \ln(1-e^{2i}) = \tan^{-1}\left(\frac{\sin 2}{1-\cos 2}\right) $$

  • $\begingroup$ Hum how do we know that the sum of e^(2i)^k/k is ln(1-e^2i)? :) Thanks for the fast response David! $\endgroup$ – majo Oct 3 '14 at 0:56
  • $\begingroup$ 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 $\endgroup$ – David Holden Oct 3 '14 at 1:19

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