Does this tricky series converge?

First examine the corresponding integral:

$$\int_2^{\infty} \frac{\cos(\log(x))}{x \log(x)}dx = \int_{log(2)}^{\infty} \frac{\cos(u)}{u} du = \left[ \frac{\sin(u)}{u} \right]_{log(2)}^{\infty} + \int_{log(2)}^{\infty} \frac{\sin(u)}{u^2}du $$ this is convergent.

Next we look at the derivative of the function $$\begin{align}f(x) &= \frac{\cos(\log(x))}{x \log(x)} \\ f'(x) &= \frac{-1/x \cdot\sin(\log(x))\cdot x \log(x) + \cos(\log(x))(\log(x)+1)}{(x \log(x))^2}\\ |f'(x)| &< \frac{2}{x^2}\end{align}$$ so we can compare the terms of sum and the integral over intervals of length 1: $$\begin{align} \left| \frac{\cos(\log(k))}{k \log(k)} - \int_k^{k+1} \frac{\cos(\log(x))}{x \log(x)} dx \right| &\leq \int_k^{k+1} | f(k) - f(x) | dx \\ &\leq \max_{k<x<k+1} | f'(x) | < \frac{2}{k^2} \end{align}$$

Since the integral and the sum of the differences both are convergent, the original series is convergent.