Criterion for a Feynman loop diagram to give a finite value

If one is dealing with renormalizable theories a rule to quickly determine whether a diagram is UV divergent or not is to ask the following question: does the vertex realized by all its external legs belong to the Lagrangian already?

Let me make this clearer: in $\lambda \phi^4$ theory in $d=4$ we only have vertices with two or four fields. So if I compute the diagram with, say, $8$ external $\phi$ legs, I am guaranteed that it will be convergent. If I instead compute a diagram with two legs that will diverge to contribute to the mass renormalization.

This is of course a bit of a silly answer because I am assuming that you renormalized the theory already, which requires the knowledge of the divergent diagrams in the first place. So how do I actually know whether a diagram converges or not?

@CStarAlgebra's answer is not wrong. You need to compute the so-called ''superficial degree of divergence.'' This is obtained by rescaling the loop momenta by a factor $\lambda$, send $\lambda \to \infty$ and look at the behavior of the integral. More precisely $$ I = \int d^Dp \,f(p,k) \;\;\mbox{has degree $A$}\;\;\Leftrightarrow\;\; \int d^Dp\,\lambda^D \,f(\lambda p,k) \underset{\lambda \to \infty}{\sim} \lambda^A I $$ When $A\geq 0$ it might signal a divergence ($A=0$ is a $\log$ divergence). But it is called superficial for a reason. There are some exceptions which sadly would require an explicit computation.

1. Gauge symmetry: Let me just make an example for this one. Take in QED a diagram with a loop of electrons and four external photons 1. This has a superficial degree that suggests it might diverge, but if it did we would need to renormalize the Lagrangian by introducing a term $(A_\mu A^\mu)^2$ which violates gauge invariance. Indeed by explicitly computing it we see that it does not diverge. And also it follows the rule of thumb I stated at the beginning.
2. Subdivergences: This applies to $>1$ loop. If we rescale all loop momenta at the same rate we are sensitive to what are usually called ''overall'' divergences. Some diagrams might instead diverge when we send to infinity only a subset of the momenta. Thus the integral might be infinite even if $A<0$. Diagrammatically this means that a subdiagram in my multi loop diagram is itself divergent. These subdivergences are totally fine because, if we renormalized correctly at the previous loop order, they should cancel identically in the final sum of all diagrams in the process. So at every loop level the new information is coming by the overall divergences.
3. IR divergences: this entire story applies to UV divergences. IR divergences appear when you have some massless degrees of freedom and the propagator diverges for soft momenta. The resolution of these kinds of divergences has nothing to do with renormalization and it is a somewhat orthogonal matter.

One can calculate the superficial degree of divergence of any diagram. This is covered in every standard introductory QFT text.