Convergence or divergence of infinite power towers of complex numbers $z^{z^{z^{z{...}}}}$

Question

Let $s$ be any complex number, $t = e^s$ and $z = t^{1/t}$. Define the sequence $(a_n)_{n\in\mathbb{N}}$ by $a_0 = z $ and $a_{n+1} = z^{a_n} $ for $n \geq 0$, that is to say $a_n$ is the sequence $z$, $z^z$, $z^{z^z}$, $z^{z^{z^{z}}}$ and so on.

I want to show that the sequence $(a_n)_{n\in\mathbb{N}}$ converges to $t$ if and only if $s$ lies in the unit disk. I know that when the sequence converges the limit is $\frac{W(-\ln(z))}{-\ln(z)}$ where $W$ is the Lambert W function.

I have verified the above statements numerically for several thousand values of $z$ but I have no idea how to actually prove it.

I graphed the natural logs of the limits on my computer. I got what appeared to be the unit disk.

If we take the log of the limit we get $\ln\left(\frac{W(−\ln(z)}{−\ln(z)}\right)=-\ln(−\ln(z))−W(−\ln(z))+\ln(−\ln(z))=−W(−\ln(z))$

So what I want to prove is equivalent to showing the sequence $a_n$ is convergent if and only if $|W(−\ln(z))| \leq 1$

I was reading the Wikipedia article on the Lambert W Function and I found this proof that the limit $c$, when it exists, is $c= \frac{W(-\ln(z))}{-\ln(z)}$

$z^c = c\implies z = c^{1/c} \implies z^{-1} = c^{-1/c} \implies 1/z = (1/c)^{1/c} \implies -\ln(z) = \frac{\ln(1/c)}{c} \implies -\ln(z) = e^{\ln(1/c)}\ln(1/c) \implies \ln(1/c) = W(-\ln(z)) \implies 1/c = e^{W(-\ln(z))} \implies \frac{1}{c} = \frac{-\ln(z)}{W(-\ln(z)} \implies c = \frac{W(-\ln(z)}{-\ln(z)}$

I can only assume that at least 1 step is not justified when $|W(-\ln(z)| > 1$ though I am not sure which one. I think that part of the problem is the equation $z^c=c$ has a solution for every non-zero complex number $c$ while the sequence $a_n$ only converges for certain special values of z. In other words the convergence of the $a_n$ is a sufficient but not necessary condition for the existence of a solution to the equation.

Answer 1

$$a_0 = z\qquad \qquad \qquad a_{n+1} = z^{a_n}$$

let $b_n = \ln a_n$ so $a_n = e^{b_n}$ and $$b_{n+1} = \ln \left(z^{e^{b_n}}\right) = e^{b_n} \ln z$$

if $b_n$ converges to $b$ then $b = e^b \ln z = - e^{b} (-\ln z)$ so

$$b = e^b \ln z = W(-\ln z)$$

where $W$ is (one of the branches of ?) the Lambert function.

let $c_n = b_n - b$ so $$c_{n+1} = e^{(b + c_n)} \ln z - b= (e^b \ln z) e^{c_n} - b = b e^{c_n} - b = b (e^{c_n} - 1)$$

suppose $|b| > 1$ and $\ln z \ne 0$. then suppose $b_n \to b$ thus $c_n \to 0$ so $e^{c_n}-1 \sim c_n$ so $c_{n+1} \sim b c_n $ which clearly cannot converge except if $c_0 = 0$ which would imply $c_n = 0$ for every $n$ which is not the case because $b_0 = \ln z$ so $b_1 = e^{\ln z} \ln z \ne b_0$ so $c_1 \ne c_0$ $\implies$ a contradiction.

hence if $\ln z \ne 0$ $$|b| = \left|W(-\ln z)\right| \le 1$$ is a necessary condition for the convergence of $(b_n)$ and $(a_n)$

Answer 2

I was reading the Wikipedia article on the Lambert W Function and I found this proof that the limit $c$, when it exists, is $c= \frac{w(-ln(z))}{-ln(z)}$

$z^c = c\implies z = c^{1/c} \implies z^{-1} = c^{-1/c} \implies 1/z = (1/c)^{1/c} \implies -ln(z) = \frac{ln(1/c)}{c} \implies -ln(z) = e^{ln(1/c)}ln(1/c) \implies ln(1/c) = W(-ln(z)) \implies 1/c = e^{W(-ln(z))} \implies \frac{1}{c} = \frac{-ln(z)}{W(-ln(z)} \implies c = \frac{W(-ln(z)}{-ln(z)}$

I can only assume that at least 1 step is not justified when $|W(-ln(z)| > 1$ though I am not sure which one.