Question
What are the convergence properties of the last equation:
$$K = e^{x} + x + \ln{x} + \ln\ln(x) + \dots $$
Can one artificially choose a value of $\ln (x)$ (since there is more than one choice of branch cuts) such that the $K$ series convergences to some series?
Background And Motivation
I was reading: Analytically continuing the series exponentially exponential series? (It is not necessary to read this post) and tried the following manipulations:
Consider the following series:
$$ S = e^x + e^{e^x} + e^{e^{e^x}} + \dots $$
We make the following observation for the $n$'th term:
Observation $1$ : $ a_n = \ln a_{n-1}$
Taking exponential both sides:
$$ e^S = e^{e^x} e^{e^{e^{x}}} \dots$$
Multiplying $e^x$ both sides we get:
$$ e^x e^S = e^x e^{e^x} e^{e^{e^{x}}} \dots$$ Now going back to the original series:
$$ S = e^x + e^{e^x} + e^{e^{e^{x}}}+ \dots $$
and now differentiating both sides:
$$ \frac{dS}{dx} = e^x + e^x e^{e^x} + e^x e^{e^x} e^{e^{e^{x}}} + \dots + e^x e^S $$
Observation $2$ : $e^xe^S = a_1 a_2 a_3 \dots $
The last term comes from observation $2$. However, we can use observation $1$ to reverse the series:
$$\frac{dS}{dx} = e^{S+x} + S+x + \ln{S+x} + \ln\ln(S+x) + \dots + e^x$$
Now we define: $J=S+x$:
$$\frac{dJ}{dx} = 1+ e^{J} + J + \ln{J} + \ln\ln(J) + \dots + e^x $$
Hence,
$$ x + \delta = \int \frac{dJ}{1+ e^{J} + J + \ln{J} + \ln\ln(J) + \dots + e^x } $$
Where $\delta$ is a constant of integration and remembering we can write $x$ as $J^{-1}(J(x))$ :
$$ J^{-1}(J(x)) + \delta = \int \frac{dJ(x)}{1+ e^{J(x)} + J(x) + \ln{J(x)} + \ln\ln(J(x)) + \dots + e^{J^{-1}(J(x))}} $$
Let, $x \to J^{-1}(x)$
$$ J^{-1}(x) + \delta = \int \frac{dx}{1+ e^{x} + x + \ln{x} + \ln\ln(x) + \dots + e^{J^{-1}(x)}} $$