I'm trying to show that $\sum \displaystyle\frac{\sin(3n\theta)}{\ln n}$ diverge or divergei know that $\displaystyle\sum \frac{1}{\ln n}$ diverge but how can i do this?

  • $\begingroup$ 1. What have you tried? 2. What is $\theta$? Do we know it? $\endgroup$ – Carl Schildkraut Oct 6 '16 at 1:05
  • $\begingroup$ You wrote "diverge or diverge" - presumably you meant "converges or diverges"? $\endgroup$ – peter a g Oct 6 '16 at 1:07

This is Dirichlet's test, $\frac{1}{\ln n}$ is decreasing to $0$ and $\sum_{n=1}^{k} \sin n\theta$ is bounded. Thus the product converges.

  • $\begingroup$ Is it obvious that that sum is bounded? Certainly the terms are but that isn't sufficient. If theta is a rational multiple of pi, it's clear, but otherwise it seems like there's an additional argument. $\endgroup$ – Richard Rast Oct 6 '16 at 1:59
  • $\begingroup$ No, a separate argument is needed to show that. There is a formula for such a sum, (easy to derive by considering the $\sum e^{in\theta}$), one then show that the sum is bounded. $\endgroup$ – Rene Schipperus Oct 6 '16 at 2:04
  • $\begingroup$ Unless theta is small, you wouldn't have convergence of that sum, either, would you? $\endgroup$ – Richard Rast Oct 6 '16 at 2:06
  • $\begingroup$ Even for small theta the sum will diverge. $\endgroup$ – Rene Schipperus Oct 6 '16 at 2:09
  • $\begingroup$ If theta is (say) -10 the sum converges by the geometric series test, but I'll admit I didn't think my comment all the way through. $\endgroup$ – Richard Rast Oct 6 '16 at 2:11

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