Series is given as
$$\sum_{m=2}^{\infty}\frac{\cos(\pi m)}{\ln(\ln(m))}$$
How does it converge?
I have shown that it diverges for the absolut series by comparison to the harmonic series. i.e $$\frac{1}{|\ln(\ln(m))|}\geq \frac{1}{n}$$ for $n\geq 2$ . Now, for convergence we use alternating series test since I can write it as $$\sum_{m=2}^{\infty}\frac{(-1)^m}{\ln(\ln(m))}$$
The problem is now occuring. I have to consider the sequence $\frac{1}{\ln(\ln(m))}$. I can not argue for monotonic decreasing since I have a asymptote at $x=e$ and thus I have to choose the biggest natural number after i.e $n=3$. Now I do not know how to continue. I can show the limit is $0$ at $\infty$.
Remember for convergence or divergence of series, all we care about is the infinite tail. We can cut off any finite starting point group of "bad" numbers and add those separately without affecting convergence/divergence.
So, you can just do the part that is $m=3+$ and chop off the bad point at $m=2$. at. Note that $\ln(m)$ is increasing throughout its domain. The composition of increasing functions is increasing, so $\ln(ln(m))$$ is increasing. Finally, the reciprocal of an increasing (positive) function is decreasing, so you get monotonic decreasing.