If a digit is written as $3.\dot{3}$, what level of infinity do the dots continue on for? Can this be proven to be true, or is it just a quirk of the notation?
More specifically, $1/3$ can obviously be broken up like
$$\sum^{\infty}_{n=0}{\frac{3}{10^n}}$$
However, I'm wondering what $\infty$ is actually meaning here. Any help is appreciated.
On
There is nothing to prove. That notations is just a convention for $$ 3.333\ldots $$ which is in turn just $$ \sum_{i=0}^\infty 3 \times 10^{-i} = 3 + \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots $$
The "$\infty$" is not a real number, nor even a cardinal number. It's part of the convention we use to write this sum of countably many real numbers, one for each natural number. You could equally well write $$ \sum_{i \in \mathbb{N}} $$ (since the order of summation does not matter for a sum of positive terms).
This means that there are countably infinite $3$'s after the one's digit. We can see this with the base-$10$ expansion of $3.\dot{3}.$ We have
$$3.\dot{3} = 3.333\dotsc = 3 + \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \dotsb = \sum_{n=0}^{\infty} \frac{3}{10^{n}}.$$