Okay, so I'm going to go out on a limb, and assume you mean that:
\[
f(z) = \exp(z) -1\\
\]
is expanded about \(z =0\), right?
I do have a long history of working with Divergent sums and Umbral Calculus and sum and series speed ups. So I'm very very happy to look into this. But I'm confused by what the bounding function is? I apologize, but I'm a tad confused.
I'd suggest an abel summation personally. And I'll explain how to do it.
Let's let \(\mathbb{E} = \{z \in \mathbb{C}\,|\, |z+\delta| < \delta\}\), so that \(0\) is on the boundary of \(\mathbb{E}\). This is just a small delta sized disk right next to \(0\), choose \(\delta\) as small as you want.
Then we know there is a holomorphic function \(g(z)\) which is holomorphic on \(\mathbb{E}\) and satisfies:
\[
g(g(z)) = e^{z} - 1\\
\]
Additionally we know that \(\lim_{z\to 0} g(z) = 0\), which is honestly just the statement that \(\eta\uparrow \uparrow z \to e\) as \(z \to \infty\), upto conjugation.
Then Abel's theorem should tell you that the power series of \(g(z)\) about \(z+\delta = 0\) is abel summable; which is confusing because it has nothing to do with how this forum uses abel functions. Abel's theorem is a very different beast. Nonetheless:
\[
\begin{align}
g(z) &= \sum_{k=0}^\infty g_k(\delta)\frac{(z+\delta)^k}{k!}\,\,\text{for}\,\, |z+\delta| < \delta\\
0 &= \lim_{z\to 0}\sum_{k=0}^\infty g_k(\delta) \frac{(z+\delta)^k}{k!}\\
\end{align}
\]
Because there is a petal about \(0\) where \(g \to 0\) which encircles \(\mathbb{E}\) for small enough \(\delta\) (We might have to finesse this a bit).
It also tells you summing the divergent series should be very doable, where as you let \(\delta \to 0\) you are approaching the coefficients you are bounding. (this is something to do with the Stolz area about the fixed point, this should definitely be doable, but Stolz has a specific condition that I'd have to reread up on to describe it).
Certainly you can use Abel sums to get a convergent/divergent series here (Again, nothing to do with the Abel function, another analysis idea by abel). It'll be different than what Norlund does, but it should work--be equivalent results. And intrinsic to that it should give a bound on the coefficients themselves. I'll try to think about this more...
This is an interesting problem, I'll have to think about it...
I apologize for all the edits, but I keep thinking of a clearer way of saying the things I'm saying. So EDIT!!!!:::::::
Thought I'd point out that you can equivalently consider \(\mathbb{E}^- = \{ z \in \mathbb{C}\,|\,|z-\delta| < \delta\}\) and the function \(g^{-1}(z)\) such that:
\[
g^{-1}(g^{-1}(z)) = \log(1+z)\\
\]
And now the Abel summation works from \(z \to 0\) from the right to left direction. Same principle though.
\[
f(z) = \exp(z) -1\\
\]
is expanded about \(z =0\), right?
I do have a long history of working with Divergent sums and Umbral Calculus and sum and series speed ups. So I'm very very happy to look into this. But I'm confused by what the bounding function is? I apologize, but I'm a tad confused.
I'd suggest an abel summation personally. And I'll explain how to do it.
Let's let \(\mathbb{E} = \{z \in \mathbb{C}\,|\, |z+\delta| < \delta\}\), so that \(0\) is on the boundary of \(\mathbb{E}\). This is just a small delta sized disk right next to \(0\), choose \(\delta\) as small as you want.
Then we know there is a holomorphic function \(g(z)\) which is holomorphic on \(\mathbb{E}\) and satisfies:
\[
g(g(z)) = e^{z} - 1\\
\]
Additionally we know that \(\lim_{z\to 0} g(z) = 0\), which is honestly just the statement that \(\eta\uparrow \uparrow z \to e\) as \(z \to \infty\), upto conjugation.
Then Abel's theorem should tell you that the power series of \(g(z)\) about \(z+\delta = 0\) is abel summable; which is confusing because it has nothing to do with how this forum uses abel functions. Abel's theorem is a very different beast. Nonetheless:
\[
\begin{align}
g(z) &= \sum_{k=0}^\infty g_k(\delta)\frac{(z+\delta)^k}{k!}\,\,\text{for}\,\, |z+\delta| < \delta\\
0 &= \lim_{z\to 0}\sum_{k=0}^\infty g_k(\delta) \frac{(z+\delta)^k}{k!}\\
\end{align}
\]
Because there is a petal about \(0\) where \(g \to 0\) which encircles \(\mathbb{E}\) for small enough \(\delta\) (We might have to finesse this a bit).
It also tells you summing the divergent series should be very doable, where as you let \(\delta \to 0\) you are approaching the coefficients you are bounding. (this is something to do with the Stolz area about the fixed point, this should definitely be doable, but Stolz has a specific condition that I'd have to reread up on to describe it).
Certainly you can use Abel sums to get a convergent/divergent series here (Again, nothing to do with the Abel function, another analysis idea by abel). It'll be different than what Norlund does, but it should work--be equivalent results. And intrinsic to that it should give a bound on the coefficients themselves. I'll try to think about this more...
This is an interesting problem, I'll have to think about it...
I apologize for all the edits, but I keep thinking of a clearer way of saying the things I'm saying. So EDIT!!!!:::::::
Thought I'd point out that you can equivalently consider \(\mathbb{E}^- = \{ z \in \mathbb{C}\,|\,|z-\delta| < \delta\}\) and the function \(g^{-1}(z)\) such that:
\[
g^{-1}(g^{-1}(z)) = \log(1+z)\\
\]
And now the Abel summation works from \(z \to 0\) from the right to left direction. Same principle though.
