I am trying to verify for which $z \in \mathbb{R}$ the series $\sum _{n=1}^{\infty } \left(1-\cos \left(\frac{1}{n}\right)\right)^z$ converges. The only test that was successful for me is the Kummer Test which gave the apparently correct result that it converges if $z > \frac{1}{2}$.
To get there I use the fact that the series $\sum _{n=1}^{\infty } a_n$ converges (when the limit exists) if $\lim_{n\to \infty } \, n \left(\frac{a_n}{a_{n+1}}-1\right) > 1$
Using Mathematica I get $\lim_{n\to \infty } \, n \left(\frac{a_n}{a_{n+1}}-1\right)=\lim_{n\to \infty } \, n \left(\left(1-\cos \left(\frac{1}{n}\right)\right)^z \left(1-\cos \left(\frac{1}{n+1}\right)\right)^{-z}-1\right) = 2z$ and therefore $z > \frac{1}{2}$.
Start reading here if you are only interested in the problem and not how i got there:
Now I try to understand how to get that
$\lim_{n\to \infty } \, n \left(\left(1-\cos \left(\frac{1}{n}\right)\right)^z \left(1-\cos \left(\frac{1}{n+1}\right)\right)^{-z}-1\right)=\lim_{x\to 0} \, \frac{-1+(1-\cos (x))^z \left(1-\cos \left(\frac{x}{1+x}\right)\right)^{-z}}{x} = 2z$
All attempts I did to show this in an elementary way failed and thats why I hope someone has an idea how to verify the result. For me its not obvious why the result to the limit is how it is