0
$\begingroup$

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? :)

|cite|improve this question
$\endgroup$
1
$\begingroup$

just a thought...

reasoning formally (i.e. leaving aside delicate questions of convergence)

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.