(07/22/2022, 02:57 PM)Gottfried Wrote: Yes, for the factorial-type series we have the Borel-summation, or even the iterated one, as K. Knopp writes it. Knopp distinguishes between Borel's integration-based method and the "simple" method (under the same name Borel).
My problem so far was that I didn't see anywhere a reliable estimate of the coefficients growthrate. so I had to explore this myself, arriving at the guess for a limiting function A(k) as mentioned in my first post in this thread - which of course were a basis for that Borel (or my experimental) method. (I don't remember why, but once in the meantime since I had the impression, the growthrate were more than hypergeometric, and thus Erdös's statement would have made sense. But I couldn't after years not reproduce the reason why I thought so).
I would like to have your (or someone's else) derivation to fix the case for sufficiency Borel-summation for some citeable statement - but as well if even it is only for the better statement in my webspace on tet.-docs... :-)
Hope you've had nice dreams - and read you at sunday -
Gottfried
Okay so first of all I want to clarify that this is for the divergent series developed to the left of \(e^{z} - 1\). This is tricky to describe but I'll give a crack at it.
We can develop a holomorphic function \(g(z)\) which is the functional square root of \(e^{z} - 1\) on the disk:
\[
\mathbb{E}_\delta = \{z \in \mathbb{C}\,|\,|z+\delta| < \delta\}\\
\]
This function is holomorphic, and within the left petal of \(e^{z} - 1\), and therefore:
\[
g(g(z)) = e^{z} -1\,\,\text{for}\,\,z \in \mathbb{E}_\delta\\
\]
Shrink \(\delta\) if necessary, but always possible.
Then:
\[
g(z) = \sum_{k=0}^\infty g_k(\delta) (z+\delta)^k\\
\]
Clearly this series diverges at \(z=0\) as it's on the boundary of \(\mathbb{E}_\delta\). But this doesn't stop us from formally rearranging the series. So... formally rearrange it:
\[
g(z) = \sum_{k=0}^\infty d_k z^k\\
\]
This series is divergent, but, under a formal manipulation:
\[
g(g(z)) = \sum_{k=1}^\infty \frac{z^k}{k!}\\
\]
This is because \(g\) satisfies this equation on \(\mathbb{E}_\delta\), and so, just as well do the coefficients. And thereby, formally rearranging the series produces the same result (we can further justify this using Abel summation).
Now, I'm going to be a tad lazy, but:
\[
g(z) = \sum_{k=0}^\infty g_k(\delta)(z+\delta)^k = \sum_{k=0}^\infty g_k(\delta)\delta^k \sum_{m=0}^k \dbinom{k}{m}\frac{z^m}{\delta^m}\\
\]
We want to write this as a power series in \(z\), trouble is, thanks to Taylor's theorem this rearrangement diverges. BUT! the rearrangements are all abel summable, because \(g(z)\) and all of its derivatives, are abel summable at \(0\).
Now we can let \(d_k \mapsto \frac{d_k}{k!}\) and let \(\delta\) be arbitrarily close to zero. Well then we get something that looks a lot like this:
\[
Bg(z) = \sum_{k=0}^\infty \frac{d_k}{k!} z^k \approx \sum_{k=0}^\infty g_k(\delta)\delta^k \sum_{m=0}^k \frac{1}{m!(k-m)!}\frac{z^m}{\delta^m}\\
\]
It'll be off by about \(O(\delta)\) for \(z\approx 0\).
Now the goal is to show that \(Bg(z)\) has a non-zero radius of convergence, and hence, Borel summable.
But, I mean, just by this approximation, you have that the right side is an entire function--this is because a holomorphic function \(f\) has an entire Borel sum.
Here's the tricky part, that I'm not certain how to pull off. I thought I could do it easier before, I'll probably have to think about it more. But if you take \(|z| < \epsilon\) for arbitrarily small and let \(\delta \to 0\). you should definitely get convergence.
I'm going to have to hunt a source to prove this though. But this is actually very very similar to the Euler case, where you are making a divergent series at a boundary point.
The function \(g(z)\) is holomorphic on a large part of the left half plane, with a divergence at \(z=0\), very much like Euler's case. And this is a general rule that these coefficients must grow like \(k!\) or \(1/\epsilon^k k!\) in our case.
Let me hunker through some textbooks, I know I've seen this before.
Another thought I had, is to just take the Euler route and just start rearranging series.
Let's write:
\[
Bg(tz) = \sum_{k=0}^\infty g_k(\delta)\frac{(tz+\delta)^k}{k!}\\
\]
This is an entire function.
Then:
\[
\lim_{\delta \to 0}\int_0^\infty e^{-t}Bg(tz)\,dt = \sum_{k=0}^\infty g_k(0)z^k\\
\]
This should converge for a significant portion of \(\Re(z) < 0\), and would be an integral representation of the Abel function. EDIT: IT CONVERGES FOR \(\Re(z) < 0\)!
I'm not going to work this out in \(\epsilon/\delta\) though, let's just stick to Euler manipulations for this, lol.
My focus was on the summation and at least now have no idea bout coefficients'.
