Background
I was recently toying with a series:
$$ S = \exp(x)+ \exp(\exp(x))+ \exp(\exp(\exp(x)))+ \dots $$
Taking exponential both sides:
$$ \implies e^S = \exp(\exp(x)) \cdot \exp(\exp(\exp(x))) \cdot \exp(\exp(\exp(\exp(x))))\cdot \dots $$
Now differentiating $ S $ with respect to $x$:
$$ \implies S' = \exp(x)+ \exp(x) \cdot \exp(\exp(x)) + \dots + \frac{e^S}{e^x} $$
Dividing both sides by $e^S$
$$ \implies S' e^{-S} = - \frac{de^{-S}}{dx}= \frac{1}{\exp(x)} + \frac{1}{\exp{(\exp(x))}} + \frac{1}{\exp(\exp(\exp(x)))} + \cdots$$
Questions
Originally $S$ does not converge, for say, $x = .01$ but the other series $ S'e^S $ does converge. Can we use this fact to analytically continue the series?(Somehow?) As I am a physics undergraduate I am aware my construction is not rigorous. How would one make a rigorous construction?
The concept of "transseries" (see Transseries for Beginners) provides a system of formal series and generalized series. However, in that formalism, the terms of a series must grow (asymptotically) smaller. Not larger, as in your example. But a variant may be considered like this: $$ T:= \exp(-x) + \exp(-\exp(x)) + \exp(-\exp(\exp(x))) + \dots $$ This one does converge (very rapidly) so it is also a series in the usual sense. And it makes sense to try your derivative computation on it.
A related object is $$ U := \frac{1}{x\cdot\log(x)\cdot\log(\log(x))\cdot\log(\log(\log(x)))\cdots} $$ which may be considered the "boundary case" between convergent integrals and divergent integrals (or series).