How can renormalised constants and perturbation expansions both be valid?

When we compute expectation values in QFT we use feynman diagrams/perturbation expansion. However, this requires the coupling constant to be small, however, when we renormalize we end up with coupling constants etc being infinite and yet we still use a perturbation expansion.

I guess a similar question is, in the path integral approach, we integrate the field over all values, but surely we will then get a situation where, for example $lambdaphi^4$, will be too big to expand?