conditional convergence of $\sum_{n=2}^{\infty} \frac{\cos(n)}{n}$ prove that the series
$$\sum_{n=2}^{\infty} \frac{\cos(n)}{n}$$
is conditionally convergent?
I tried to prove that it is not absolutely convergent series by trying to prove that $\sum_{n=2}^{\infty} \frac{\vert\cos(n)\vert}{n}$ is divergent, but I did not succeed. Is there another way?

Dirichlet's test does the job pretty nicely. Let we prove that the partial sums of $\cos n$ are bounded:
$$ \left|\sum_{n=1}^{N}\cos(n)\right| = \left|\frac{\sin\left(N+\frac{1}{2}\right)-\sin\left(\frac{1}{2}\right)}{2\sin\left(\frac{1}{2}\right)}\right|\leq\frac{1+\sin\frac{1}{2}}{2\sin\frac{1}{2}}<\frac{14}{9} $$
and we are done, since $\left\{\frac{1}{n}\right\}_{n\geq 1}$ is a sequence monotonically decreasing to zero.
We may even compute the value of the series: by summation by parts,

$$\sum_{n\geq 1}\frac{\cos n}{n}=\text{Re}\left(-\log(1-e^i)\right)=\color{red}{-\log\left(2\sin\frac{1}{2}\right)}=0.0420195\ldots $$

On the other hand, the series $\sum_{n\geq 1}\frac{\left|\cos n\right|}{n}$ is clearly divergent by a simple combinatorial lemma: given three consecutive integers $n,n+1,n+2$, for at least one of them the value of $\left|\cos(\cdot)\right|$ is greater than $\sin(1)$, hence:
$$ \sum_{n=1}^{3N}\frac{\left|\cos n\right|}{n}\geq \sum_{k=1}^{N}\frac{\sin(1)}{3k} $$
and the RHS is divergent as $N\to +\infty$.
