Consider the series


where $0<\theta_0<\pi$. Is there one easy way to prove that it diverges? I wanted to used this to prove that the series $\sum_n \frac{\sin|n\theta_0|}{n}$ diverges, so comparing with this series wouldn't be an option here.

I tried the ratio test, but didn'g get far, I needed to evaluate the limit

$$\lim_{n\to \infty}\dfrac{1-\cos(2(n+1)\theta_0)}{2(n+1)}\dfrac{2n}{1-\cos(2n\theta_0)},$$

but this also seems complicated. Is there one easier way to show that this series diverges? How can I prove this?

  • $\begingroup$ with Abel: divergent$-$ convergent$=$divergent. $\endgroup$ – hamam_Abdallah Oct 24 '16 at 19:24
  • $\begingroup$ @AbdallahHammam Does $$\sum_n \frac{cos(2n\theta_0)}{2n}$$ actually converge ? A moment ... Do you mean the corresponding improper integral ? $\endgroup$ – Peter Oct 24 '16 at 19:26
  • $\begingroup$ This series actually converges as its similar to the alternating harmonic series, essentially $\sin(|n\theta_0|) "=" (-1)^n$ (more or less). The alternating series test will show that it converges. $\endgroup$ – Triatticus Oct 24 '16 at 19:30
  • 1
    $\begingroup$ Have a look at en.wikipedia.org/wiki/Dirichlet%27s_test and math.stackexchange.com/questions/17966/… $\endgroup$ – Jack D'Aurizio Oct 24 '16 at 19:43
  • $\begingroup$ @Peter Yes, by Abel's rule, like the improper integral. $\endgroup$ – hamam_Abdallah Oct 24 '16 at 19:45

Your Answer

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

Browse other questions tagged or ask your own question.