Convergence of $\sum_{n=1}^{\infty}\left(\left(\sqrt{n}\right)\left(n^{\alpha\cos(\frac{1}{n})}-n-\cos\left(\frac{1}{n}\right)\right)\right)^{-1}$

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Convergence of the following series as $\alpha \in \mathbb{R}$

$$\sum_{n=1}^{\infty}\left(\left(\sqrt{n}\right)\left(n^{\alpha\cos(\frac{1}{n})}-n-\cos\left(\frac{1}{n}\right)\right)\right)^{-1}$$

As $n \to +\infty$ we have $a_n\sim \left(n^{\frac{1}{2}}\left(n^{\alpha}-n-1\right)\right)^{-1} = \left(n^{\frac{2\alpha+1}{2}}-n^\frac{3}{2}-n^\frac{1}{2}\right)^{-1}$

Hence, as $0<\alpha<1$ we have $a_n =\mathcal{O}\left(\frac{1}{n^\frac{3}{2}}\right)$, as $n \to +\infty$, that converges.

As $\alpha>1$ we have $a_n=\mathcal{O}\left(\frac{1}{n^\frac{2\alpha+1} {2}}\right)$, as $n \to +\infty$, which converges in turn.

Is it right or I got rid of too much informations in the asymptotic expansion?

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We have that

$$\cos\left(\frac{1}{n}\right)=1+O\left(\frac1{n^2}\right)$$

$$n^{\alpha\cos(\frac{1}{n})}=e^{\alpha\cos(\frac{1}{n})\log n}=e^{\alpha\log n+O\left(\frac{\log n}{n^2}\right)}=n^{\alpha}\left(1+O\left(\frac{\log n}{n^2}\right)\right)=n^{\alpha}+O\left(\frac{\log n}{n^{2-\alpha}}\right)$$

$$n^{\alpha\cos(\frac{1}{n})}-n-\cos\left(\frac{1}{n}\right)=n^{\alpha}+O\left(\frac{\log n}{n^{2-\alpha}}\right)-n-1+O\left(\frac1{n^2}\right)$$

and therefore

  • for $\alpha<1$

$$\left[\sqrt{n}\left(n^{\alpha\cos(\frac{1}{n})}-n-\cos\left(\frac{1}{n}\right)\right)\right]^{-1} \sim-\frac1{n\sqrt n}$$

  • for $\alpha=1$

$$\left[\sqrt{n}\left(n^{\alpha\cos(\frac{1}{n})}-n-\cos\left(\frac{1}{n}\right)\right)\right]^{-1} \sim-\frac1{\sqrt n}$$

  • for $\alpha>1$

$$\left[\sqrt{n}\left(n^{\alpha\cos(\frac{1}{n})}-n-\cos\left(\frac{1}{n}\right)\right)\right]^{-1} \sim \frac1{n^{\alpha+\frac12}}$$

and the given series converges if and only if $\alpha \neq 1$.