Does the series $\sum_{n=2}^{\infty} \frac{\cos(\ln(\ln(n)))}{\ln(n)}$ converge? Could someone please give a hint about how to solve the convergence of the following series:
$$\sum_{n=2}^{\infty} \frac{\cos(\ln(\ln(n)))}{\ln(n)}$$
 A: The following is not a rigorous proof, but more of an intuitive proof with some gross estimations relying on the fact that the logarithm ascends very slowly. I hope it still answers the question. I believe this can be made into a rigorous proof with some care for details.
First consider absolute convergence: $|\cos \left({\ln \ln n }\right)|$ spends most of its time being between let's say $0.1$ and $1$, so usually the absolute value is bigger than $\frac {0.1} {\ln n}$, and the sum diverges. (You can replace $0.1$ with some fixed $\epsilon$.)
For conditional convergence the situation is more delicate. The only thing you can hope for is that the sign changes will cause cancellation between the positive and the negative parts (like in Leibniz's series test). So let's consider this possibility.
The cosine changes its sign in increments of $\pi$. So suppose the sign changes at some $n$, and the very next time it changes is at some $m > n$. Then we have
$$\ln \ln m \approx \pi + \ln \ln n,$$
which means
$$\ln m \approx C \ln n,$$
where $C = e^\pi \approx 23.14$, which finally means
$$m \approx n^C.$$
The next-next time the sign changes is at $k \approx m^C \approx n^{C^2}$.
The first sign-invariant part of the sum (between $m$ and $k$), in absolute value, again supposing that most of the time the cosine is between $0.1$ and $1$, is bigger than... (apologies for gross estimations...) $$\sum_{j=n}^{n^C} \frac {0.1} {\ln j} \approx \int_n^{n^C} \frac {0.1}{\ln x}\,dx \approx 0.1 \left.\frac x {\ln x}\right|_n^{n^C} \approx \frac {0.1 n^C}{C \ln n} $$
The second sign-invariant part of the sum will similarly be $$\frac {{0.1}n^{C^2}}{C^2 \ln n}$$
The ratio between the second and the first is $$\frac 1 C n^{C^2 - C}$$
This means that the summand spends way too much time being of constant sign, so much that the value being added now is much, much bigger than the amount that was subtracted last time. The series therefore fluctuates between very positive and very negative values, and diverges.
A: Let 
$$
S_l=\sum_{k\in E_l}\frac{\cos(\log(\log(n)))}{\log(n)}
$$
where $$E_l=\{k\in \mathbb{N}: \exp\bigl(\exp(2\pi l-\frac{\pi}{3})\bigr)\le k\le \exp\bigl(\exp(2\pi l)\bigr)\}$$
So we have,
$\#(E_l)>\exp\bigl(\exp(2\pi l)\bigr)-\exp\bigl(\exp(2\pi l-\frac{\pi}{3})\bigr)-1=\exp\bigl(\exp(2\pi l)\bigr)-\exp\bigl(\alpha\exp(2\pi l)\bigr)-1
$
where $0<\alpha=\exp(\frac{-\pi}{3})<1$.
Then for all $k\in E_l$, we have $\frac{\cos(\log(\log(k)))}{\log(k)}\ge \frac{\cos(\frac{-\pi}{3})}{\exp(2\pi l)}=\frac{1}{2\exp(2\pi l)}$
Thus $S_l\ge \frac{\exp\bigl(\exp(2\pi l)\bigr)-\exp\bigl(\exp(2\pi l-\frac{\pi}{3})\bigr)-1}{2\exp(2\pi l)}$
To conclude notice that $\lim_{x\rightarrow +\infty}\frac{\exp(x)-\exp(\alpha x)-1}{2x}=+\infty$
Your series diverges.
