Convergence of the following series as $\alpha \in \mathbb{R}$
$$\sum_{n=1}^{\infty}\left(\left(\sqrt{n}\right)\left(n^{\cos(\frac{1}{n^{\alpha}})}-n-\cos\left(\frac{1}{n}\right)\right)\right)^{-1}$$
Some days ago I posted the same series but the parameter was in another position.
Since, as $n \to \infty$ we have $\cos(\frac{1}{n^{\alpha}}) \sim 1-\frac{1}{2n^{2\alpha}}$, how to interpret the series?
If $\alpha=0$ then $$\sqrt n\,\Bigl(n^{\cos 1}-n-\cos\frac1n\Bigr)\sim-n^{3/2}$$ and the series converges. If $\alpha\ge0$ then \begin{align} n^{\cos n^{-\alpha}}-n&=n\bigl(n^{(\cos n^{-\alpha})-1}-1\bigr)\\ &=n\bigl(e^{((\cos n^{-\alpha})-1)\log n}-1\bigr)\\ &\sim n\,(\cos n^{-\alpha}-1)\log n\\ &\sim-\frac{\log n}{2\,n^{2\alpha-1}} \end{align} If $\alpha>1/2$ then $$ n^{\cos n^{-\alpha}}-n-\cos\frac1n\sim-1 $$ and the series diverges. If $\alpha=1/2$ then $$ n^{\cos n^{-\alpha}}-n-\cos\frac1n\sim-\frac12\log n $$ and the series diverges. Finally, if $0\le\alpha<1/2$ then $$ \sqrt n\,\Bigl(n^{\cos n^{-\alpha}}-n-\cos\frac1n\Bigr)\sim-\frac{\log n}2n^{3/2-2\alpha}; $$ the series converges if $0<\alpha<1/4$ and diverges if $\alpha\ge1/4$.
When $\alpha<0$ things are more complicated, and the answer depends on how close to one $\cos n^{-\alpha}$ can get.