Let $\{ z_n \} \subset \mathbb{C}$ $z_n \neq 0$. I'm trying to prove that convergence of $\Pi z_n$ is equivalent to:
$\forall \epsilon>0$ $\exists $ $\nu>0$ such that if $n>\nu$ $|z_{n+1}······z_{n+k}-1|<\epsilon$ with $k \in \mathbb{N}$
Since $\mathbb{C}$ is a complete convergence equals Cauchy-Sequence. My problem starts when I try to prove that the statement from above implies the sequence of partial products is Cauchy.
How can i show that (with the property from above) $\exists K>0$ such that $\forall N \in \mathbb{N}$ $\quad$$| \prod_{n=1}^{N} z_n|<K?$
Choosing $\epsilon = \frac{1}{2}$ in the property, we have the existence of a $\nu$ such that
$$\left\lvert 1 - \prod_{m=1}^k z_{n+m}\right\rvert < \frac{1}{2}$$
for all $n > \nu$ and $k\in \mathbb{N}\setminus \{0\}$. In particular,
$$\frac{1}{2} < \left\lvert \prod_{m=1}^k z_{n+m}\right\rvert < \frac{3}{2}$$
for all $n > \nu$ and $k\in \mathbb{N}$.
Then it follows that
$$\frac{1}{2} \min \left\{ \left\lvert \prod_{n=1}^k z_n\right\rvert : k \leqslant \nu+1 \right\} \leqslant \left\lvert \prod_{n=1}^N z_n\right\rvert \leqslant \frac{3}{2} \max \left\{ \left\lvert \prod_{n=1}^k z_n\right\rvert : k \leqslant \nu+1\right\},$$
which shows that all partial products are bounded away from $0$ and from $\infty$. Hence a further application of the property shows that the sequence of partial products is a Cauchy sequence.