A way to prove convergence and also to understand why this integral represents potential energy between 2 charge distributions. Another way to look at this is
$$\iiint \epsilon_0\vec{E_1}.\vec{E_2}d\tau$$ $$\iiint \epsilon_0\vec{E_1}.(-\nabla V_{2})d\tau$$
Using the vector field identity
$$\nabla \cdot (V_{2}\vec{E_1}) = V_{2} \nabla \cdot \vec{E_1} + \vec{E_{1}} \cdot \nabla V_{2}$$
$$\vec{E_{1}} \cdot (-\nabla V_{2})= -\nabla\cdot(V_{2}\vec{E_1}) + V_{2} \nabla \cdot\vec{E_1} $$
Substituting this in the integral gives
$$\iiint\epsilon_0( -\nabla \cdot (V_{2}\vec{E_1}) + V_{2} \nabla \cdot \vec{E_1} )d\tau$$
$$\iiint-\epsilon_0 \nabla \cdot (V_{2}\vec{E_1}) d\tau+\iiint \epsilon_0V_{2} \nabla \cdot \vec{E_1} d\tau$$
Envoking stokes theorem on the first integral
$$\iint-\epsilon_0 (V_{2}\vec{E_1}) \cdot d\vec{S}+ \iiint\epsilon_0 V_{2} \nabla \cdot \vec{E_1} d\tau$$
Lim V goes to infinity the surface integral evaluates to zero
$$\iiint\epsilon_0V_{2} \nabla \cdot \vec{E_1} d\tau$$
Envoking gauss law
$$\iiint\epsilon_0 V_{2} \frac{\rho_1}{\epsilon_0} d\tau$$
$$\iiint V_{2} \rho_1 d\tau$$
In this form it is clear to see why it represents potential energy. As you are building up charge distribution 1 in the presence of potential 2
In this form you can also understand why it converges, as the V$\rho$ is always a finite number as V is never evaluated at any point that causes it to be infinity