# New/useful method for summation of divergent series?

## 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:

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

• Reason for downvote? – More Anonymous Jul 1 '18 at 13:46