The usual QCD folklore tells us that perturbative expansions are (at best) asymptotic. Recently a colleague of mine told me that the expansion of the beta function is thought to be convergent because, and I quote, "it only sees the UV".
This would-be argument does not convince me at all but is he right? and if so, why?
The beta function of a generic theory is typically expected to be asymptotic too. (Note the "generic": some theories with extra symmetry, e.g., enough supersymmetry, have convergent or even vanishing beta functions).
For example, by instanton considerations, the beta function of $\phi^4$ theory is expected to be $\beta_n(g)\sim(-1)^nn!\,n^{7/2}\times\text{subleading}$. Indeed, the explicit loop computation yields (cf. 1807.05060) $$ \beta(g)=3g^2-5.7g^3+32.5g^4-271.6g^5+2848.6g^6-34776g^7+474651g^8+\mathcal O(g^9) $$ which roughly follows the estimate above, up to one order of magnitude.
Of course, $\phi^4$ is not asymptotically free so it is not clear whether the expression above is meaningful at all. But it does illustrate the general behaviour of typical beta functions.
The divergence of the beta function does not contradict the fact that it may be chosen to be two-loop exact by a suitable renormalisation scheme, cf. this PSE post.
The beta function is scheme dependent beyond two loops. I suspect that it should be possible to devise a scheme (which is useless in practice) in which all coefficients of the beta function vanish beyond two loops. In supersymmetric gauge theories without matter the exact perturbative beta function was determined by NSVZ, http://inspirehep.net/record/192068?ln=en, (in a more useful scheme). The NSVZ beta function has a perfectly convergent perturbative expansion (except for poles).
