Depend on what text you read, an infinite product with an infinite number of terms that are 0 is either divergent, or diverge to 0. Even though, obviously, the partial product is still a convergence sequence.
However, an infinite sum with an infinite number of term that are 0 do not subject to this treatment. Given the precedence that with infinite product, why do people not define infinite sum in such a way to make special exception to series with infinite number of terms that are 0?
Whether to say that an infinite product with $\prod_{n=1}^N a_n\to 0$ "converges to $0$" or "diverges to $0$" is a matter of terminology. There are good reasons to prefer the second, which have to do with tests for convergence/divergence. Products that tend to $0$ do not exhibit the behavior we normally associate with convergence. For example, their terms may be unbounded, like $$0\cdot 1\cdot 2\cdot 3 \cdot \dots = 0$$ or, to give a less obvious example, $$1\cdot \frac{1}{1^2}\cdot 2\cdot \frac{1}{2^2}\cdot 3\cdot \frac{1}{3^2} \cdot\dots =0$$ If we accept these products as convergent, all tests for divergence go out of the window.
Another reason is that under the application of an exponential function, convergent series precisely correspond to infinite products with nonzero limit.
Finally, from the viewpoint of applications, for functions that are naturally represented as infinite products it turns out to be important to know whether they are zero or not. Having the formula $\zeta(s) = \prod_{p}(1-p^{-s})^{-1}$ is nice for many reasons, but in part because it tells us that $\zeta(s)\ne 0$ for $\operatorname{Re}s>1$.