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$$\cfrac{z}{1+z} \cdot \cfrac{z}{1+z-\cfrac{z}{1+z}} \cdot \cfrac{z}{1+z-\cfrac{z}{1+z-\cfrac{z}{1+z}}} \cdots= 1-\frac{1}{z}$$

I propose that this works for any $z \in C$ if and only if $|z|>1$.

That much I got by experimenting and by the identity I already proved using Euler's continued fraction formula:

$$\cfrac{z}{1+z-\cfrac{z}{1+z-\cfrac{z}{1+z-\cfrac{z}{1+z-\cdots}}}}=\begin{cases}1 & |z| > 1\\z & |z| \leq 1\end{cases}$$

But I don't know how to prove the above product. Any ideas?

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Well, let's start by calling the terms of your product $a_n$, that is, $a_1=\frac{z}{1+z}, a_{n+1}=\frac{z}{1+z-a_n}$. Some experimenting suggests the formula:

$$a_n=\frac{z+z^2+...+z^n}{1+z+z^2+...+z^n}=\frac{z(1-z^n)}{1-z^{n+1}}$$

and this can be verified fairly easily by induction. Now, let's consider the partial products:

$$a_1a_2\cdots a_n=\frac{z(1-z^1)}{1-z^{2}}\cdot \frac{z(1-z^2)}{1-z^{3}}\cdots \frac{z(1-z^n)}{1-z^{n+1}}=z^n\frac{1-z}{1-z^{n+1}}=\frac{z-1}{z-z^{-n}}$$

Considering the cases $|z|<1, |z|>1$ separately, we see that:

  • $|z|<1$ causes the product to "diverge to 0"
  • $|z|>1$ lets the product converge to $\frac{z-1}{z}$
  • $|z|=1$ has varied behaviour - if the $z$ is a root of unity, the product will contain terms of both $0$ and $\infty$, and if not, it will vary without converging.
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  • $\begingroup$ But numerically I get $1-\frac{1}{z^2}$, not $1-\frac{1}{z}$, $\endgroup$ – Yuriy S Feb 22 '16 at 15:41
  • $\begingroup$ Perhaps because $1-\frac{1}{z^2}=\frac{1-\frac{1}{z}}{\left(\frac{z}{1+z}\right)}$ - might be missing the first term? $\endgroup$ – πr8 Feb 22 '16 at 15:43
  • $\begingroup$ You are right, my algorithm was wrong, thank you $\endgroup$ – Yuriy S Feb 22 '16 at 15:45

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