New/useful method for summation of divergent series?

Question

Questions

$$ S(n,x) = x+e^x + e^{e^x} + e^{e^{e^x}} + \dots \text{$n$ times}$$

Also obeys (see background for argument):

$$ \frac{1}{2 \pi i} \oint e^{S(k,x)} \frac{\partial \ln(\frac{\int_0^\infty e^{-t} t^k dt }{ \int_0^\infty e^{-t} t^{(k-n)} dt})}{\partial k} dk = \frac{\partial S(n,x)}{\partial x}$$

Can this be used in the Borel summation sense for divergent series? If so, when can it be used for analytical continuity (convergence issues)? Is it useful(intuitively I feel it should be more powerful than Borel summation)? In the following heuristic sense:

$$ \kappa = \sum_{n=1}^\infty a_n = \sum_n a_n \frac{\frac{\partial S(n,x)}{\partial x}}{1 + e^x + e^x e^{e^x} + e^x e^{e^x} e^{e^{e^x}} + \dots \text{$n$ times} } $$

Using the L.H.S of the first equation:

$$ \kappa = \frac{1}{2 \pi i} \sum_n a_n \frac{\oint e^{S(k,x)} \frac{\partial \ln(\frac{\int_0^\infty e^{-t} t^k dt }{ \int_0^\infty e^{-t} t^{(k-n)} dt})}{\partial k} dk }{1 + e^x + e^x e^{e^x} + e^x e^{e^x} e^{e^{e^x}} + \dots \text{$n$ times} } $$

Swapping order of summation and contour integral:

$$ \kappa =^! \frac{1}{2 \pi i} \oint e^{S(k,x)} \sum_n \frac{ a_n \frac{\partial \ln(\frac{\int_0^\infty e^{-t} t^k dt }{ \int_0^\infty e^{-t} t^{(k-n)} dt})}{\partial k} }{1 + e^x + e^x e^{e^x} + e^x e^{e^x} e^{e^{e^x}} + \dots \text{$n$ times} }dk $$

How can I make this rigorous?

Background

I've been recently studying the following series:

$$ S(n,x) = x+e^x + e^{e^x} + e^{e^{e^x}} + \dots \text{$n$ times}$$

Where the $n$'th term is raising the $x$ exponentially $n$ number of times.

$$ b_n(x) = \underbrace{e^{e^{e^{\dots}x}}}_{\text{$n$ times exponentially raised}} $$

$n$ number of times.

Hence, we notice:

$$ e^{S(r,x)} = \frac{\partial b_{r+1}(x)}{\partial x}$$

Summing both sides and defining $S(0,x) \equiv 0$:

$$ \sum_{r=0}^n e^{S(r,x)} = \sum_{r=1}^{n+1} \frac{\partial b_{r}(x)}{\partial x} $$

Hence, we get:

$$ \sum_{r=0}^n e^{S(r,x)} = \frac{\partial S(n+1,x)}{\partial x}$$

Rewriting the R.H.S using complex analysis as a contour integral over the whole complex plane:

$$\frac{1}{2 \pi i} \oint \sum_{r=0}^n \frac{e^{S(k,x)}}{k-r}dk = \frac{\partial S(n+1,x)}{\partial x}$$

Taking $e^{S(k,x)}$ common:

$$ \frac{1}{2 \pi i} \oint e^{S(k,x)} \sum_{r=0}^n \frac{1}{k-r}dk = \frac{\partial S(n+1,x)}{\partial x}$$

Further using $d \ln x = dx/x$

$$ \frac{1}{2 \pi i} \oint e^{S(k,x)} \sum_{r=0}^n d \ln({k-r}) = \frac{\partial S(n+1,x)}{\partial x}$$

Rewriting as factorial:

$$ \frac{1}{2 \pi i} \oint e^{S(k,x)} \frac{\partial \ln(\frac{(k)!}{(k-n-1)!})}{\partial k} dk = \frac{\partial S(n+1,x)}{\partial x}$$

Analytically continuing $k!$ using the gamma function:

$$ \frac{1}{2 \pi i} \oint e^{S(k,x)} \frac{\partial \ln(\frac{\int_0^\infty e^{-t} t^k dt }{ \int_0^\infty e^{-t} t^{(k-n-1)} dt})}{\partial k} dk = \frac{\partial S(n+1,x)}{\partial x}$$

Answer 1

(This is no answer but continuation of my earlier comments, the answer-box is used because of the picture that I want to show).

Here is a small image on the existence and some relations of period-2 -points, in the hope to get an idea how to find period-2 points of alternate signs, following the rule:
$$e^{p_1} \to p_2 ; e^{p_2} \to p_1; e^{p_1} \to p_2; \cdots \\ S(p_1) = p_1 + p_2 + p_1 + p_2 + \cdots = ? $$

Observations: there are "a lot of" (infinitely many) pairs of 2-periodic points. They occur even in subsets, each which can be described as sequence of pairs (here colored with different colours, pairs of one subset in the same color).
The subsets with color yellow, green and blue have sequences of pairs which converge to 1-periodic (fixed-) points!

But I still didn't find pairs which alternating signs in the real components...