The series of functions $\{f_n\}$ is given by $$f_n=\sum_{k=0}^{n}\frac{2^k}{z^k+z^{-k}} \quad \text{for } |z|<\frac{1}{2}.$$
The exercise given to me is to show that $\{f_n\}$ converges uniformly.
My attempt:
\begin{align} |f_n-f_m|\leq |\sum_{k=m}^{n}\frac{2^k}{z^k+z^{-k}}| \leq\sum_{k=m}^{n}\frac{2^k}{|z^k+z^{-k}|} \end{align}
Therefore, \begin{align} \|f_n-f_m\| \leq \Big\| \sum_{k=m}^{n}\frac{2^k}{|z^k+z^{-k}|} \Big\|. \end{align} The sum is biggest for $|z^k+z^{-k}|$ smallest. If we consider the function $g(z)=z^k+z^{-k}$ we see that it is holomorphic on $D(0,\frac{1}{2})$ except for $z=0$. Thus the maximum has to occur on the boundary. Moreover $|g|$ has a lower bound $=\frac{1}{2^k}+2^k$. However, doing this gets us \begin{align} \|f_n-f_m\| \leq \sum_{k=m}^{n}\frac{2^k}{2^k+2^{-k}} . \end{align}
which is obviously not Cauchy for $k,m>N$ for sufficiently large $N$.
My Question: Are the series of functions uniformly convergent and if yes can I have a hint as how to approach this?