<s>[INST] Prove that the function $f(z) = \sum_{n=0}^\infty \dfrac{1}{1+z^n}, \quad |z|>1,$ is holomorphic. I attempted to solve the following problem and would like feedback (corrections, suggestions for improvement, or a better way).
Prove that the series $\sum_{n=0}^{\infty} \dfrac{1}{1+z^n}$ converges for $|z|>1$ and that the function 
$$f(z) = \sum_{n=0}^\infty \dfrac{1}{1+z^n}, \quad |z|>1,$$
is holomorphic.
\textbf{Solution:} 
For $|z|>1$, we note that  $|z^n| -1  \leq |1+ z^n| \leq |z^n| + 1 $ which implies 
$$\dfrac{1}{|z|^n -1}  \geq \dfrac{1}{|1+ z^n|} \geq \dfrac{1}{|z|^n + 1}.$$
Letting $r=|z|$, we note that by the comparison test for series,
\begin{eqnarray}
\lim_{n \to \infty} \dfrac{\dfrac{1}{r^{n+1}+1}}{\dfrac{1}{r^n+1}} &=& \lim_{n \to \infty} \dfrac{r^n+1}{r^{n+1}+1} \\
&=&  \lim_{n \to \infty} \dfrac{r^n}{r^{n+1}} \\
&=& \dfrac{1}{r} \\
&<&1, \  \text{since} \  r>1,
\end{eqnarray}
that is converges.  Also,
\begin{eqnarray}
\lim_{n \to \infty} \dfrac{\dfrac{1}{r^{n+1}-1}}{\dfrac{1}{r^n-1}} &=& \lim_{n \to \infty} \dfrac{r^n-1}{r^{n+1}-1} \\
&=&  \lim_{n \to \infty} \dfrac{r^n}{r^{n+1}} \\
&=& \dfrac{1}{r} \\
&<&1, \  \text{since} \  r>1,
\end{eqnarray}
that is converges.  So by the Sandwich" orSqueeze" Theorem",  $$\sum_{n=1}^\infty\dfrac{1}{|1+ z^n|}$$
converges. Hence $\sum_{n=1}^\infty\dfrac{1}{1+ z^n}$ converges since it converges absolutely for $|z|>1$. 
To show that $f(z)$ is holomorphic we can either show that $f(z)$ is differentiable in the region $G$ and apply Goursat's theorem or we can show that $f(z)$ is continuous in $G$ and for every closed path $C$ in $G$, $\int_C f(z) \ dz = 0$.
I'm having difficulty showing $f(z)$ is point-wise continuous.  For every $\varepsilon > 0$, there exists at $\delta > 0$ such that
$$|f(z)-f(z_0)| < \varepsilon, \quad \text{ if } \ |z-z_0| < \delta.$$
\begin{eqnarray}
|f(z)-f(z_0)| &=& \left| \sum_{n=0}^\infty \dfrac{1}{1+z^n} - \sum_{n=0}^\infty  \frac{1}{1+z_0^n}\right|  \nonumber \\
&=& \left| \sum_{n=0}^\infty \dfrac{1}{1+z^n} - \frac{1}{1+z_0^n}\right|  \nonumber \\
&=& \left| \sum_{n=0}^\infty \dfrac{z_0^n-z^n}{(1+z^n)(1+z_0^n)}\right|  \nonumber \\
&\leq& \sum_{n=0}^\infty  \left|\dfrac{z_0^n-z^n}{(1+z^n)(1+z_0^n)}\right|  \nonumber \\
&=& \sum_{n=0}^\infty  \left|\dfrac{(z_0-z)(z_0^{n-1}+z_0^{n-2}z+z_0^{n-3}z^2+ \cdots + z_0z^{n-2}   +z^{n-1})}{(1+z^n)(1+z_0^n)}\right|  \nonumber \\
&=& \sum_{n=0}^\infty  |z_0-z| \cdot \left|\dfrac{(z_0^{n-1}+z_0^{n-2}z+z_0^{n-3}z^2+ \cdots + z_0z^{n-2}   +z^{n-1})}{(1+z^n)(1+z_0^n)}\right| \nonumber \\
&<& \sum_{n=0}^\infty  \delta \cdot \left|\dfrac{(z_0^{n-1}+z_0^{n-2}z+z_0^{n-3}z^2+ \cdots + z_0z^{n-2}   +z^{n-1})}{(1+z^n)(1+z_0^n)}\right|  \nonumber \\
&=&  \delta \cdot\sum_{n=0}^\infty  \left|\dfrac{(z_0^{n-1}+z_0^{n-2}z+z_0^{n-3}z^2+ \cdots + z_0z^{n-2}   +z^{n-1})}{(1+z^n)(1+z_0^n)}\right|  \nonumber \\
&\leq &  \delta \cdot\sum_{n=0}^\infty  \dfrac{|z_0^{n-1}|+|z_0^{n-2}z|+|z_0^{n-3}z^2|+ \cdots +| z_0z^{n-2}| + |z^{n-1}|}{(|z^n|-1)(|z_0^n|-1|)}  \nonumber \\
&= &  \delta \cdot\sum_{n=0}^\infty  \dfrac{|z_0^{n-1}|+|z_0^{n-2}||z|+|z_0^{n-3}||z^2|+ \cdots +| z_0||z^{n-2}| + |z^{n-1}|}{(|z^n|-1)(|z_0^n|-1|)} \nonumber \\
&\leq &  \delta \cdot\sum_{n=0}^\infty  \dfrac{|z_0^{n-1}|+|z_0^{n-2}|(|z_0|+\delta) +|z_0^{n-3}|(|z_0|+\delta)^2+ \cdots +| z_0|(|z_0|+\delta)^{n-2} + (|z_0|+\delta)^{n-1}}{((|z_0|+\delta)^{n}-1)(|z_0^n|-1|)}  \nonumber \\
&< &  \delta \cdot\sum_{n=0}^\infty  \dfrac{2n \cdot (|z_0|+\delta)^{n-1}}{(|z_0|^n-1|)^2}   \nonumber \\
&=& \delta \cdot M \nonumber \\
&<& \varepsilon
\end{eqnarray}
I call $M$ the sum to which the series converges (by the series ratio test).   Is there a better way to show that $f(z)$ is point-wise continuous?
To use Morera's theorem, we need to show uniform convergence of $f(z)$.  But I ran into problems with that as well.
Thanks for your assistance.
 [/INST] This is hard work! Take $R>1$ and consider $U=\{z:|z|>R\}$. On $U$,
$$\left|\frac{1}{1+z^n}\right|\le\frac{1}{R^n-1}\le\frac{R^{-n}}{1-1/R}$$
(for $n\ge1$). So on $U$ the series is uniformly convergent, and the sum
is therefore holomorphic on $U$.
</s>