This is a question that has bothered myself and Gottfried Helms a fair amount of late. He has made his case for the following result, but a proof escapes both of us. The question is deceptively simple, but keeps eluding each of my attempts when we get into the finer details.

Let's start by calling:

$$ f(z) = e^{z}-1\\ $$

Then $f$ has a parabolic fixed point at $0$ with multiplier $1$. By which we have:

$$ f(z) = z + O(z^2)\\ $$

There exists an attracting petal, and a repelling petal; and conveniently they are separated by $i\mathbb{R}$. By which, we are only focused on the attracting petal, which is given as $\Re(z) < 0$. Using the standard Ecalle construction of an Abel function, which I'm mostly just following Milnor's *Dynamics in One Complex Variable*. We can construct an Abel function:

$$ \alpha(f(z)) = \alpha(z) + 1\\ $$

This function is $2 \pi i$ periodic, and is holomorphic for $\Re(z) < 0$. We are primarily focused near the boundary point $z = 0$--where the Abel function has a singularity.

Now we can invert this to the left of zero, so that we can construct a super function $\alpha^{-1}(z)$, and by this, we can get to our function in question:

$$ g(z) = \alpha^{-1}(\alpha(z) + 1/2)\\ $$

I will spare the details, but this function is also $2 \pi i$ periodic, and holomorphic for $\Re(z) < 0$. We don't need this though, just that it's holomorphic on a domain to the left of $0$, with $0$ on the boundary. This function satisfies the equation:

$$ g(g(z)) = f(z)\\ $$

Gottfried first came to me with this question, and so I am not the most versed in the construction of this series, but it relates to a question which he had investigated on this forum--of the $\sin$ half iterate. But, the essential statement, is that we can construct a formal series:

$$ h(z) = \sum_{n=1}^\infty d_n z^n\\ $$

Such that this series satisfies the formal manipulations:

$$ h(h(z)) = \sum_{n=1}^\infty \frac{z^n}{n!}\\ $$

By which this series has a radius 0 convergence (it's a divergent series).

The question then becomes, is it possible to represent this divergent series, as Euler would represent the divergent series:

$$ \sum_{n=0}^\infty (-1)^nn!z^n\\ $$

Using the Laplace transform:

$$ \int_0^\infty \frac{e^{-zx}}{1+x}dx = \sum_{n=0}^\infty (-1)^nn!z^n\\ $$

Or, rather, could we do something similar, in the same vein that Borel summation is done.

This means, we have the function $g(z)$, which is holomorphic, and a boundary value at $0$. We have a divergent/formal series at $z=0$. If we were to take:

$$ \mathcal{B}h(z) = \sum_{n=0}^\infty d_n\frac{z^n}{n!}\\ $$

Will this have a non-zero radius of convergence? By which we could Borel sum the divergent series. I know I'm missing something crucial, and every attempt I've made a using traditional arguments fails because $g(z)$ is a rather anomalous function.

Essentially, can we make the same treatment of Euler's series, of this series. By which we'd be able to pull out a bound $d_n = O(c^n n!)$ for some $c > 0$.