# Regularization of Exponentially exponential series?

## 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, $x \to J^{-1}(x)$

$$J^{-1}(x) + \delta = \int \frac{dx}{1+ e^{x} + x + \ln{x} + \ln\ln(x) + \dots + e^{J^{-1}(x)}}$$

• K should converge in the upper half of the complex plane; but it has complex singularities at 0,1,e,e^e,e^e^e ... etc. – Sheldon L Jun 1 '18 at 6:56
• actually, the ln^[n](x) will have a limiting value of the upper fixed point of ~= 0.318+1.337i, so the sum will be unbounded. – Sheldon L Jun 1 '18 at 16:09