Is there an easy way to prove that this series diverges?

Asked Oct 24 '16 at 19:15

Active May 11 '20 at 09:58

Viewed 108 times

Consider the series

$$\sum_{n}\dfrac{1-\cos(2n\theta_0)}{2n},$$

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?

edited May 11 '20 at 09:30

Martin Sleziak

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asked Oct 24 '16 at 19:15

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with Abel: divergent$-$ convergent$=$divergent. – hamam_Abdallah Oct 24 '16 at 19:24
@AbdallahHammam Does $$\sum_n \frac{cos(2n\theta_0)}{2n}$$ actually converge ? A moment ... Do you mean the corresponding improper integral ? – Peter Oct 24 '16 at 19:26
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. – Triatticus Oct 24 '16 at 19:30
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Have a look at https://en.wikipedia.org/wiki/Dirichlet%27s_test and http://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro – Jack D'Aurizio Oct 24 '16 at 19:43
@Peter Yes, by Abel's rule, like the improper integral. – hamam_Abdallah Oct 24 '16 at 19:45
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Does this answer your question [use $\sin^2(t)=1/2(1-\cos(2t))$]? Convergence of $\sum_{n=1}^\infty \frac{\sin^2(n)}{n}$ – Integrand Sep 07 '20 at 14:50

Is there an easy way to prove that this series diverges?

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