How to prove the convergence of the serie $\sum \frac{\sin(3n\theta)}{\ln n}$?

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?

• 1. What have you tried? 2. What is $\theta$? Do we know it? – Carl Schildkraut Oct 6 '16 at 1:05
• You wrote "diverge or diverge" - presumably you meant "converges or diverges"? – 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.
• 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. – Rene Schipperus Oct 6 '16 at 2:04