Nature of $\sum\left(\cos \frac{1}{n^\alpha}\right)^n$ 
How can I study the convergence of this series  $$\sum\cos^n
\frac{1}{n^\alpha}$$ depending on $\alpha >0$
  ?

I found that $$\cos^n \frac{1}{n^\alpha} \sim \exp\left(\frac{-1}{2n^{2\alpha -1}}\right)$$
The case $2\alpha - 1  \geq 0$ can then be treated, as the series diverges.
But how can I exploit that same similar to treat the other case?
Thanks
 A: Using two facts, this one and this one, we have
$$1-\frac{x^2}{2}\leq \cos{x}\leq e^{-\frac{x^2}{2}}, x \in \left[0,\frac{\pi}{2}\right]$$
or, because $\frac{1}{n^{\alpha}}$ will be close to $0$ for suficiently large $n$
$$1-\frac{1}{2n^{2\alpha}}\leq
\cos{\frac{1}{n^{\alpha}}}\leq
\frac{1}{e^{\frac{1}{2n^{2\alpha}}}}$$
and, applying Bernoulli's inequality
$$1-\frac{n}{2n^{2\alpha}}\leq
\left(1-\frac{1}{2n^{2\alpha}}\right)^n\leq
\cos^n{\frac{1}{n^{\alpha}}}\leq
\frac{1}{e^{\frac{n}{2n^{2\alpha}}}}$$
Thus, for $2\alpha-1\geq0$
$$\lim\limits_{n\to\infty}\cos^n{\frac{1}{n^{\alpha}}} \ne 0$$
and the series doesn't converge.

For $2\alpha-1<0$ or $0<1-2\alpha$ and
$$0<\cos^n{\frac{1}{n^{\alpha}}}\leq
\frac{1}{e^{\frac{n}{2n^{2\alpha}}}}=
\left(\frac{1}{e}\right)^{\frac{n^{(1-2\alpha)}}{2}}$$
Using this limit (more details here)
$$\lim\limits_{n\to\infty}\frac{n^{1-2\alpha}}{\ln{n}}=\infty$$
for suficiently large $n$ we will have 
$$n^{1-2\alpha}>4\ln{n}=\ln{n^4} \Rightarrow \\
e^{\frac{n^{1-2\alpha}}{2}}>e^{\frac{\ln{n^4}}{2}}=n^2 \Rightarrow \\
\left(\frac{1}{e}\right)^{\frac{n^{1-2\alpha}}{2}}<\frac{1}{n^2}$$
and by comparison test, the series converges.
A: Note that the expression we are dealing with now is 
$$\sum_{n=1}^\infty \exp(-n^a/2)$$
with $0<a<1$. Now we can use the integral test and a substitution $u^{1/a}=x$, $\frac{u^{1/a-1}}{a}du=dx$
$$\int_1^\infty \exp(-x^a/2)dx = \frac{1}{a}\int_1^\infty \exp(-u/2)u^{1/a-1}du$$
And the last integral converges via the ratio test.
