Absolute convergence of series $\sum_{n=1}^\infty \frac{1}{\frac{1}{z^n}-1}$

Question

$$ \begin{align} \sum_{n=1}^\infty \frac{1}{\frac{1}{z^n}-1} \end{align} $$

Determine the values of $z,z\in\mathbb{C}$ so that the series converges absolutely

I know that the series converges for $|z|<1$.

If a series $\sum_{n=0}^\infty \alpha_n $ converges , then $\alpha_n \to 0 , n\to \infty$, which in this case indicates

that $\frac{1}{z^n}-1 \to \infty , n \to \infty$, and that leads to $z^n<1$, which means that $|z|<1$

My problem is that is my solution formal enough? and actually I'm not sure if it guarantees absolute convergence.

Answer 1

First of all, $|z|<1$ is necessary, but maybe not sufficient (harmonic series). But here it is:

\begin{align} \sum_{n=1}^\infty\left|\dfrac{1}{\frac{1}{z^n}-1}\right|=\\ \sum_{n=1}^\infty\left|\dfrac{z^n}{1-z^n}\right|\leq\\ \tag{1}\sum_{n=1}^\infty\dfrac{|z|^n}{1-|z|}=\\ \frac{1}{1-|z|}\sum_{n=1}^\infty|z|^n=\\ \left(\frac{1}{1-|z|}\right)^2<\infty\\ \end{align}

To arrive at $(1)$ we used the reverse triangle inequality for all $n\geq1$:$$\left|1-z^n\right|\geq1-|z|^n\geq1-|z|$$

So for $|z|<1$ the given series converges absolutely. Your solution is wrong, in the sense that convergence implies $a_n\to 0$, but $a_n\to 0$ does not imply convergence