Does $\sum_{k=2}^{\infty}\frac{\cos(k)}{\log(\log(k))}$ converge?

Question

Does $\sum_{k=2}^{\infty}\frac{\cos(k)}{\log(\log(k))}$ converge?

According to Wolfy, it seems to converge to about -8.80.

This is suggested by Determine convergence of a series

This could possibly be done by looking at the $k$ for which $m < \log(\log(k)) < m+1 $ (or $e^{e^m} < k < e^{e^{m+1}}$) and looking at $\dfrac1{m}\sum\cos(k)$ in this region and showing that it is small enough so that the sum of these converges.

The fact that $\sum_{k=1}^m \cos(k)$ is bounded would probably be useful, but I don't think that this is enough.

Answer 1

By Dirichlet's test and the fact $\sum\limits_{k=1}^n \cos(k)$ is bounded, $\frac{1}{\log\log(k)} \to 0$ is monotone, we're done.

Answer 2

You can probably use Diriclet's test here if I am not wrong, as the sum of cos(k) is bounded and 1/log(log(k)) is monotone and approaches 0.

Answer 3

Hint: Use the following $$\sum_{k=1}^n \cos k = \frac{2}{\sin 1}\sum_{k=1}^{n} \sin(k+1)-\sin (k-1)$$ to prove the boundedness of the sum of $\cos$'s.