Does $\sum\limits_{n=1}^{\infty}\frac{\cos^{2}(n+1)}{n}$ converge?

Regarding the question on absolute convergence, the answer is no.

$$\sum_{n=1}^m\frac{\cos^2 (n+1)}{n} = \underbrace{\sum_{n=1}^m\frac{1}{2n}}_{\text{divergent harmonic series}} + \underbrace{\sum_{n=1}^m\frac{\cos 2(n+1)}{2n}}_{\text{convergent by Dirichlet test}}$$

Also we can use the same approach to prove convergence when $(-1)^n$ appears.

$$\sum_{n=1}^m\frac{(-1)^n\cos^2 (n+1)}{n} = \underbrace{\sum_{n=1}^m\frac{(-1)^n}{2n}}_{\text{convergent alternating series}} - \underbrace{\sum_{n=1}^m\frac{(-1)^{n+1}\cos 2(n+1)}{2n}}_{\text{convergent by Dirichlet test}}$$

Note that $(-1)^{n+1}\cos 2(n+1) = \cos [(n+1)\pi] \cos 2(n+1) = \cos [(n+1)(2 + \pi)]$

$\sum_{n=1}^{\infty}\frac{\cos^{2}(n+1)}{n}$ clearly diverges. For any two consecutive integers, at least one of them is distant from a (non-integer) half-integer multiple of $\pi$ by at least $\frac{1}{2}$, which is more than $\frac{\pi}{8}$. That number of the pair will have a cosine greater than $\cos(\frac{3\pi}{8})$ in absolute value, and $\cos(\frac{3\pi}{8})=\sin(\frac{\pi}{8})>\frac{1}{2}\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{4}>\frac{1}{3}$. Thus, any two consecutive terms $\frac{\cos^{2}(j+1)}{j} + \frac{\cos^{2}((j+1)+1)}{j+1}$ will contribute at least $\frac{1}{9(j+1)}$ to the sum. Since $j\ge 1$, we have $\frac{1}{j+1}\ge\frac{1}{4j}+\frac{1}{4(j+1)}$, so $\frac{1}{9(j+1)} \ge\frac{1}{36j}+\frac{1}{36(j+1)}$, i.e. $$\frac{\cos^{2}(j+1)}{j} + \frac{\cos^{2}((j+1)+1)}{j+1} \ge \frac{1}{36j}+\frac{1}{36(j+1)}$$ But this means $$\sum_{n=1}^{\infty}\frac{\cos^{2}(n+1)}{n}\ge\frac{1}{36}\sum_{n=1}^{\infty}\frac{1}{n}$$ which diverges.

You could use Dirichlet's test. Observe $n^{-1}$ converges monotonically to zero and \begin{align} \left|\sum^N_{n=1}(-1)^n \cos^2(n+1) \right|\leq 1. \end{align} Hence the series converges.