Limit of geometric series sum when $r = 1$

In order to understand why $r=1$ is not allowed you have to look at the proof of the geometric series (I will neglect the constant $a$). We start with

$$S_n = 1 + r+ r^2+ ... + r^n$$ $$rS_n = r + r^2 + r^3 +... + r^{n+1}$$

Then we subtract both equations.

$$S_n ( 1-r) = 1 - r^{n+1}$$

Solving for $S_n$ requires that $r\neq 1$ or we would dived by $0$.

What happens is that the equality$$\sum_{k=0}^nar^n=\frac{a-ar^{n+1}}{1-r}$$only holds when $r\neq1$. When $r=1$, it doesn't make sense. So, in order to study the behaviour of the series $\displaystyle\sum_{k=0}^nar^n$ when $r=1$, we have to take another apprach. And that approach is:$$\sum_{k=0}^na1^n=\sum_{k=0}^na=(n+1)a.$$

It is a matter of terminology. You are right to say that "its behaviour is DEFINED", but this is an informal statement, meaning that we understand that the series diverges to infinity.

But when the limit of a series cannot be expressed by a real number, we say that it is undefined. You must accept this convention.