I think both the link and the question refer to Dyson's heuristical argument on why the perturbative series in QED could not be convergent. It goes somewhat like this:

Suppose the series in $\alpha$ converges in some radius. The it converges also for negative values of the coupling constant inside that radius. Consider now what kind of theory is QED with a negative $\alpha$. In that theory like charges attract and opposite sign charges repel each other. Now take the vacuum of the non-interacting theory. This state is unstable against formation of electron-positron pairs, because said pairs would repel indefinitely leading t a lower energy state. You can make an even lower energy state by adding pairs that would separate in two clusters of electrons on one side and positrons on the other. Therefore this theory does not possess a ground state, since the spectrum is unbounded from below. Hence there is no consistent QED for negative coupling constant. And so the perturbative series cannot converge.

As far as I know this argument is strictly heuristical, but shortly after it appeared (in the 1950s) Walter Thirring proved the divergence for a particular scattering process (I'm not in my office so I don't have the correct reference, but I'm positive the paper is in Thirring's selected works as well as explained in his autobiography).

Note that this question of convergence was proeminent in a period where people tried to define QFT in terms of the perturbative expansion. The advent of non-perturbative effects (instantons, confinements, pick your favorite...) coupled with renormalization group showed that this was the wrong approach for QFT.

But note also that the argument of vacuum instability depends on the interaction. It does not preclude the possibility of designing a QFT with convergent perturbative expansion, it just shows that it it not to be expected in a general theory.

This post imported from StackExchange Physics at 2017-10-16 12:26 (UTC), posted by SE-user cesaruliana