Here's my try on this one:
Put $f(z)=\cos({\frac{1}{z-1}})$ as a complex series around $0$. Find the convergence radius.
Here's what I did so far:
$\cos({\frac{1}{z-1}})=\sum_{n=0}^{\infty}(-1)^n\cdot\frac{\displaystyle\frac{1}{(z-1)^{2n}}}{(2n)!}=\sum_{n=0}^{\infty}(-1)^n\cdot\frac{1}{(z-1)^{2n}(2n)!}=\sum_{n=0}^{\infty}(-1)^n\cdot\frac{1}{(w)^{2n}(2n)!}=\sum_{n=0}^{\infty}(-1)^n\cdot\frac{1}{(t)^{-n}(2n)!},$
with $z-1=w$ and $w^{-2}=t$.
So from here, we find the radius of convergence:
$\frac{1}{R}=\lim_{n\to\infty}{\frac{|a_{n+1}|}{|a_n|}}=\lim_{n\to\infty}{\frac{(-1)^{n+1}(2n)!}{(2n+2)!\cdot (-1)^n}}=0\implies R=\infty.$
Is this right? Maybe I missed something or did something wrong and I can't see it.
Thanks for your time.