I don't understand Dyson's argument for divergence of perturbative QED

Question

Dyson's argument that the perturbative expansion for QED must diverge:

[...] let $$F(e^2)=a_0+a_1e^2+a_2e^4+\ldots$$ be a physical quantity which is calculated as a formal power series in $e^2$ by integrating the equations of motion of the theory over a finite or infinite time. Suppose, if possible, that the series... converges for some positive value of $e^2$; this implies that $F(e^2)$ is an analytic function of $e$ at $e=0$. Then for sufficiently small value of $e$, $F(−e^2)$ will also be a well-behaved analytic function with a convergent power series expansion[...]

He then argues that for negative coupling $-e^2$ (like-charges attract) the vacuum of positive-coupling QED will be unstable and will continuously generate positron-electron pairs of a lower energy state (that subsequently repel). That part of the argument is fine.

The issue I have is with the argument that $F(e^2)$ must be analytic at $e^2=0$. What's to prevent $F(e^2)$ being well-behaved on $e^2\gt 0$ and poorly-behaved on $e^2\lt 0$. E.g: $$ F(e^2) = \exp(e^2), e^2\gt 0\\ F(e^2) = \frac{1}{e^2}, e^2\lt 0 $$ and we make no claim for $F(0)$?

The Taylor series for the exponential converges everywhere, so the perturbative expansion of $F(e^2)$ is convergent for $e^2\gt 0$, as we would like to be true of QED. But this function does not have a convergent perturbative expansion in an open set around zero.

But nothing about the lack of a convergent perturbative expansion of $F(e^2)$ around $e^2=0$ implies that there is not a convergent perturbative expansion of $F(e^2)$ for $e^2\gt 0$. What am I missing?

(It seems to me that Dyson's argument is simply an argument that $F(e^2)$ must have such behavior around $e^2=0$. Switching from opposite-charges attract to opposite-charges repel while keeping matter/anti-matter pair creation the same (i.e. anti-matter has the opposite charge to matter) is not remotely a "continuous" change, no matter how small the coupling constant).

Answer 1

$F(z) = a_0 + a_1 z + \ldots$ is a power series in $z$ centered at $z=0$. If this series converges for some $z_0>0$, then its radius of convergence is at least $z_0$ and so it is guaranteed to converge on the open ball $|z|<z_0$.

If I understand your objection, you're saying that it's possible to have a function which is non-analytic at $0$ but which is equal to $F(z)=a_0 + a_1 z + \ldots\ $ for some (or possibly all) $z>0$. That is of course true, as your example demonstrates, but you're getting the argument twisted. The function which Dyson is referring to is the power series in question, defined on the domain on which it converges. In other words, letting $F$ be a formal power series, we are interested in the function $f:U\rightarrow \mathbb R$ where $$U:= \{z\in \mathbb R \ | \ F(z) \text{ converges}\}$$ $$f :U\ni z\mapsto F(z)$$

It is this $f$ which Dyson is referring to as being analytic at $0$ if it converges for some $z_0>0$. In your objection, you are saying that one could construct some $\hat f$ which coincides with $f$ for $z>0$ but is non-analytic at $0$, which is beside the point.

Answer 2

By being analytic at $e=0$ means that it has a perturbation expansion with a non-zero radius of convergence when expanded about $e=0$. As the radius of convergence is determined by the distance to the nearest singularity, means that there can be no nearby singularity for either positive or negative $e$. But there is is such a singularity (a branch point exactly at $e=0$) so the radius of convergence is zero, and by definition the function is not analytic at $e=0$.