I would like to understand the problem of the convergence/divergence of the series appearing in QED: my question is inspired by a great article of John Baez, which can be found online. On the page 23 of this article we read:
,,However, if we continue adding up terms in this power series, there is no guarantee that the answer converges. Indeed, in 1952 Freeman Dyson gave a heuristic argument that makes physicists expect that the series diverges, along with most other power series in QED. The argument goes as follows. If these power series converged for small positive $\alpha$, they would have a nonzero radius of convergence, so they would also converge for small negative $\alpha$. ''

I don't understand this argument: $\alpha$ is certainly small but the radius of convergence may be still smaller and we don't run into the problem of convergence of the series for negative $\alpha$.

What is the reason to believe that the radius of convergence would be greater that the value of $\alpha$?

Also I would like to ask:

If it is widely believed that the series actually diverges what is the general belief about the nature of this divergence: do the corrections tends to $0$ but too slowly to ensure convergence, or do they blow up or maybe do some other crazy things (like oscillating)? If it the case for which term it is believed to start happening?

  • $\begingroup$ There is a theory behind this: asymptotic series. Even though we cannot compute things arbitrarily precisely, we do know the result up to some error that can be controlled. $\endgroup$
    – user178876
    Dec 18, 2018 at 23:22
  • $\begingroup$ A related topic: physics.stackexchange.com/questions/422975/… $\endgroup$
    – maplemaple
    Dec 19, 2018 at 3:47
  • $\begingroup$ Why do you think there is no convergence problem for negative $\alpha$? The fact that there is an "obvious" clumping instability if alpha takes any negative value is part of the reason for believing that the series has radius of convergence 0. $\endgroup$
    – Buzz
    Dec 19, 2018 at 3:56


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.