Why is the sum to infinity of a geometric distribution equal to 1? I know that the geometric distribution follows the rules of a geometric progression thus by using the sum to infinity formula (which I know its proof and is really convinced by it),
$$\frac {a}{1-r}$$
We can easily arrive at this:
$\frac {p}{1-(1-p)} =\frac {p}{1-1+p} = \frac pp = 1$
But the thing is, when my teacher asked my class what we would expect the sum to be no one hesitated to say 1. They all knew it by logic. However I couldn't understand how they arrived at that.
One guessing was that maybe the sum is 1 because "P", probability of getting my success at first trial will be added to small fractions of it, but I still couldn't arrive at anything, after all, these fractions are going to keep decreasing to very tiny numbers so I couldn't get my head around the fact that they'll end up as the same value as P and thus their sum will be 1.
Can someone explain to me the logic or intuition behind this?

They are probably using the fact that the sum of the probabilities of all possible outcomes must equal $1$. This is true for any [discrete] random variable.
They are probably not doing the computation $p + p(1-p) + p(1-p)^2 + \cdots = \frac{p}{1-(1-p)} = 1$ in their heads instantaneously.
