In QFT is said that the renormalized Dyson series is only asymptotic. But to be able to say it is necessary to know to what function of $g$ (the coupling constant) the Dyson series is asymptotic.
For example, suppose that some transition amplitude $A(g)$ is given perturbatively by a series of powers of $g$. In order to prove that this series is asymptotic to $A(g)$ I need to know the value of $A(g)$ non-perturbatively, but this is not possible since the only way $A(g)$ is given is the Dyson series.
Best Answer
Suppose that $A(g)$ is given by some perturbative expansion around $g=0$: $$A(g) = \sum_n c_n g^n.$$ Then the statement that this expansion is asymptotic means that the radius of convergence is zero: for fixed $g$, no matter how small, the limit $$\lim_{N \to \infty} \; \sum_{n < N} c_n g^n$$ diverges. Typically it's enough to know that large-$n$ behaviour of the coefficients $c_n$ to establish this. In QFT, sometimes you can estimate the magnitude of the $c_n$ (e.g. you count the number of Feynman diagrams at $n$-loop order multiplied by the typical contribution of a single diagram). For instance, if $$c_n \sim n!$$ you can prove that the series is asymptotic. You do this using the tools you learned in undergrad calculus (for example the ratio test).
Notice that this is a really fast growing series: if for instance $c_n \sim a^n$ for some constant $a$, then you have convergence inside a disk $|g| \leq 1/a$. So you need even faster than exponential growth of the $c_n$ to have an asymptotic series.