TITLE: Solving a tough limit
QUESTION [5 upvotes]: 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

REPLY [2 votes]: The key is to use the Taylor series approximation $$\cos x \approx 1 - \frac{x^2}{2},$$ which is good for small $x$. That applies to both $x$ and $x/(1+x)$, and so $$\frac{1-\cos x}{1-\cos \frac{x}{1+x}} \approx (1+x)^2.$$ Substituting this into your expression, we get that the expression is $$\approx \frac{(1+x)^{2z}-1}{x} \approx \frac{(1+2zx)-1}{x} = 2z.$$ This reasoning can be made formal using big-O (or little-O) notation, but I leave that for you to ponder.