Convergence or divergence of $\sum_{n=1}^{\infty}\frac{1}{n^{2-\cos n}}$ 
Is the below infinite sum convergent or divergent?
  $$\sum_{n=1}^{\infty}\frac{1}{n^{2-\cos n}}$$

I considered $1\le2-\cos(n) \le 3$ but I am not able to determine whether this series is convergent or divergent.
I know that this series is similar to $p$-series but the sequence consists of case $p=1$ and $p\gt1$.
Any help would be appreciated.
 A: Here it is a translation of the article of Brighi mentioned in the comments. All credits to him.

The problem of understanding if the series 
$$ \sum_{n\geq 1}\frac{1}{n^{2+\cos n}}, \qquad \sum_{n\geq 1}\frac{1}{n^{2-\cos n}} $$
are convergent is far from being simple. It is well known that the outcome depends on the distribution of $\cos(n)$ in $[-1,1]$. Our proof of the divergence of such series exploits different lemmas from Analysis and Number Theory.

It is practical to recall that for any $\alpha\in\mathbb{R}^+\setminus\mathbb{Q}$ the convergents $\frac{p_n}{q_n}$ of the continued fraction of $\alpha$ fulfill $\left|p_n-\alpha q_n\right|<\frac{1}{q_n}$. In the opposite direction we have Lagrange's theorem, stating that $\left|p-\alpha q\right|<\frac{1}{2q}$ ensures that $\frac{p}{q}$ belongs to the convergents of the continued fraction of $\alpha$.

If $f\in C^1([a,b])$, for any $x\in[a,b]$ we have
$$ \left|(b-a)f(x)-\int_{a}^{b}f(t)\,dt\right|\leq (b-a)\int_{a}^{b}\left|f'(t)\right|\,dt.$$
It is sufficient to partition $[a,b]$ as $[a,x]\cup[x,b]$, apply integration by parts to $\int_{a}^{x}$ and $\int_{x}^{b}$, then apply the triangle inequality.
The Denjoy-Koksma inequality: Let $z\in\mathbb{R}$ and let $f:\mathbb{R}\to\mathbb{R}$ a $1$-periodic function of class $C^1$. Given $\alpha\in\mathbb{R}^+\setminus\mathbb{Q}$, let $\frac{p}{q}$ be a convergent of the continued fraction of  $\alpha$. For any integer $n\in\{0,1,\ldots,q-1\}$, let $x_n=n\alpha-\lfloor n\alpha\rfloor$. We have:
$$ \left|\int_{0}^{1}f(t)\,dt -\frac{1}{q}\sum_{n=0}^{q-1}f(z+x_n)\right|\leq\frac{1}{q}\int_{0}^{1}|f'(t)|\,dt.$$
Proof: by the periodicity of $f$ we may assume without loss of generality that $z=0$ and that $\frac{p}{q}<\alpha<\frac{p}{q}+\frac{1}{q^2}$. Since $\alpha$ is irrational all the points $x_0,x_1,\ldots,x_{q-1}$ are distinct. Let $\ell\in\{1,2,\ldots,q\}$. Since $p$ and $q$ are coprime, there are some $n\in\{0,1,\ldots,q-1\}$ and some $k\in\mathbb{Z}$ such that $np+kq=\ell-1$. By considering
$$ 0< n\alpha-\frac{np}{q}<\frac{1}{q},\qquad \frac{\ell-1}{q}\leq n\alpha+k<\frac{l}{q} $$
we have that $k=-\lfloor n\alpha\rfloor$. Since the points $x_0,x_1,\ldots,x_{q-1}$ are distinct, for any $\ell\in\{1,2,\ldots,q\}$ there is a unique integer $n_\ell\in\{1,2,\ldots,q\}$ such that
$$ n_\ell \in\left[\frac{\ell-1}{q},\frac{\ell}{q}\right).$$
By exploiting the lemma before the claim we get:
$$\begin{eqnarray*}\left|\int_{0}^{1}f(t)\,dt-\frac{1}{q}\sum_{n=0}^{q-1}f(x_n)\right|&\leq&\sum_{\ell=1}^{q}\left|\int_{(\ell-1)/q}^{\ell/q}f(t)\,dt-\frac{1}{q}f(x_{n_\ell})\right|\\&\leq &\frac{1}{q}\sum_{\ell=1}^{q}\int_{(\ell-1)/q}^{\ell/q}|f'(t)|\,dt = \frac{1}{q}\int_{0}^{1}|f'(t)|\,dt.\end{eqnarray*}$$

Let $a\geq 1$ and $q\geq 2$ be natural numbers. We trivially have
$$ \sum_{n=(a-1)q+1}^{aq}\frac{1}{n^{2+\cos(n+\theta)}}\geq \sum_{n=(a-1)q+1}^{aq}\frac{1}{(aq)^{2+\cos(n+\theta)}}=\frac{1}{(aq)^2}\sum_{n=(a-1)q+1}^{aq}(aq)^{-\cos(n+\theta)}. $$
On the other hand
$$ \int_{0}^{1}(aq)^{-\cos(2\pi t+\theta)}\,dt\geq (aq)\int_{-\frac{\theta}{2\pi}-\frac{1}{2}}^{-\frac{\theta}{2\pi}+\frac{1}{2}}e^{-\frac{1}{2}\log(aq)(2\pi t+\theta+\pi)^2}\,dt $$
such that
$$ \int_{0}^{1}(aq)^{-\cos(2\pi t+\theta)}\,dt\geq \frac{aq}{2\pi\sqrt{\log(aq)}}\int_{0}^{2\pi\sqrt{\log(aq)}}e^{-s^2/2}\,ds \geq C_0\frac{aq}{\sqrt{\log(aq)}}.$$
Now we may consider the convergents $\frac{p_k}{q_k}$ of the continued fraction of $\alpha=\frac{1}{2\pi}$. By defining $x_n=n\alpha-\lfloor n\alpha\rfloor$, $z_{a,k}=(a-1)q_k+1$ and $f_{a,k}=(aq_k)^{-\cos(2\pi x+\theta)}$ we may apply the Denjoy-Koksma inequality to $f_{a,k}$, picking $q=q_k$ and $z=z_{a,k}$. This leads to
$$ \left|\int_{0}^{1}f_{a,k}(t)\,dt -\frac{1}{q_k}\sum_{n=0}^{q_k-1}f(z_{a,k}x+x_n)\right|\leq 2a$$
then to
$$ \sum_{n=(a-1)q_k+1}^{aq_k}\frac{1}{n^{2\cos(n+\theta)}}\geq \frac{C_0}{a\sqrt{\log(aq_k)}}-\frac{2}{aq_k}.$$
By exploiting $H_n\sim\log(n)$ we easily get
$$ \sum_{n=1}^{q_k^2}\frac{1}{n^{2+\cos(n+\theta)}}\geq C_1\sqrt{\log q_k}-\frac{\log q_k}{k} $$
and by letting $k\to +\infty$ the divergence of $\sum_{n\geq 1}\frac{1}{n^{2+\cos(n+\theta)}}$ is proved.

The original problem is the instance $\theta=\pi$.
