Trying to evaluate $$\ln(\ln(\ln(\ln(\cdots\ln(x)\cdots))))$$For some fixed $x$ produces a complex answer that appears to converge, at least sometimes.
So I want a proof that this converges for either some $x$, no $x$, or all $x$.
If it converges for all $x$ or some $x$, what does it converge to?
If it diverges, is there a way we can evaluate it like we evaluate diverging sums?
And after all of that, does it appear to converge to the same value, no matter what $x$ value we start with?
I know $\ln(z)=\ln(|z|)+i\arg(z)$, but I can't repeat this process without a given $z$. (where $z$ is complex).
A similar post of mine found here does not answer my question and focuses more on the limits, calculus, and infinites.
This question asks for consideration from a complex-analysis point of view, considering convergence of value in the complex plane.
I can't answer your question for a complex number, but if $x\in\Bbb R^+$ you can only iterate $\log$ finitely many times before it becomes complex.
Let $b_0=e$ and $b_{n+1}=e^{b_n}$. Let $x\in[1,\infty)$, $a_0=x$, $a_{n+1}=\log a_n$, and let $N$ be the largest integer such that $b_N\le x<b_{N+1}$.
So, $$b_N\le a_0<b_{N+1}$$.
Iterating $\log$ $N$ times on this inequality gives: $$b_0\le a_N<b_1$$. In other words, $$\begin{align} e\le a_N<e^e & \implies 1\le a_{N+1}<e\\ & \implies 0\le a_{N+2}<1. \end{align}$$
So, $a_{N+3}$ will be negative, and the next iteration will be complex. If you're taking $\log$ as a real-valued function, you can only iterate it finitely many times before it becomes undefined.