Applying the ratio test and uniform convergence

Question

I have been trying to apply the ratio test onto $\dfrac{z^n}{1+z^n}$. After the usual initial steps.

I need to show that $$\lim_{n \to \infty} \left|\dfrac{z(1+z^n)}{1+z^{n+1}}\right|<1$$

I am unsure of how to make further progress, and so what is the trick from here?

Once this is shown, we can therefore say that the series $$\displaystyle\sum\limits^\infty_{n=0} \dfrac{z^n}{1+z^n}$$ converges absolutely.

Before I can deduce $\displaystyle\sum\limits^\infty_{n=0} \dfrac{z^n}{1+z^n}$ is holomorphic on $D=\{z \in \mathbb{C} \mid |z| <1\},$ would I have to show uniform convergence? If so, what is the simplest way?

Answer 1

For $\;|z|<1\;$ , we have that

$$\lim_{n\to\infty}\left|z\frac{1+z^{n+1}}{1+z^n}\right|\le\lim_{n\to\infty}\,|z|\,\frac{1+|z|^{n+1}}{1-|z|^n}=|z|\frac{1+0}{1-0}=|z|<1$$

Answer 2

For $z_{0}\in D$, we choose a small $\eta>0$ such that $|z_{0}|+\eta<1$ and consider the ball $B_{\eta}(z_{0})\subseteq D$. We have $|1+z^{n}|\geq 1-|z|^{n}\geq 1-(|z-z_{0}|+|z_{0}|)^{n}\geq 1-(|z_{0}|+\eta)^{n}\geq 1-(|z_{0}|+\eta)>0$ for all $z\in B_{\eta}(z_{0})$.

Now we see that \begin{align*} \left|\dfrac{z^{n}}{1+z^{n}}\right|\leq\dfrac{|z|^{n}}{1-(|z_{0}|+\eta)}\leq\dfrac{(|z_{0}|+\eta)^{n}}{1-(|z_{0}|+\eta)} \end{align*} for all $z\in B_{\eta}(z_{0})$. Use M-Test to conclude the uniform convergence.

For the holomorphicity, we first look at the partial sum of the derivatives, which is $\displaystyle\sum_{n=1}^{N}\dfrac{nz^{n-1}}{(1+z^{n})^{2}}$. Still, this can be controlled by $\left|\dfrac{nz^{n-1}}{(1+z^{n})^{2}}\right|\leq\dfrac{n|z|^{n-1}}{(1-(|z_{0}|+\eta))^{2}}\leq\dfrac{n(|z_{0}|+\eta)^{n-1}}{(1-(|z_{0}|+\eta))^{2}}$, once again use M-Test to conclude. As a result, the function is holomorphic at $z=z_{0}$. Since $z_{0}$ is arbitrary in $D$, the function is holomorphic throughout.