# What level of infinity is referred to when talking about recurring digits?

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.

• The symbol $\infty$ in this notation reminds of the extra point in the one-point compactification of $\Bbb N$ Jun 22, 2019 at 17:02

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

$$3.\dot{3} = 3.333\dotsc = 3 + \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \dotsb = \sum_{n=0}^{\infty} \frac{3}{10^{n}}.$$

• So it could also be that there is $3.333\dots 3$? I mean, a sequence of length $\Bbb N\cup\{\Bbb N\}$ is countably infinite... Jun 22, 2019 at 17:04
• @AsafKaragila wouldn't that mean that it would be possible to rewrite $3.\dot{3}$ as $$\sum^{|\mathbb{N}|}_{n=0}{\frac{3}{10^n}}$$ in which case $$3\times\sum^{|\mathbb{N}|}_{n=0}{\frac{3}{10^n}}$$ would be a countable infinite sequence, so $$10 = 2\times\sum^{|\mathbb{N}|}_{n=0}{\frac{3}{10^n}}+\sum^{|\mathbb{N}|+1}_{n=0}{\frac{3}{10^n}}$$, yet since Euler's proof states that $9.\dot{9}=10$, $1=0$ at the top of the sum function in the last expression. This is the main reason I was confused about this in the first place, because this disproof is obviously wrong, yet so simply to follow. Jun 22, 2019 at 18:29
• @Geza: My point is that sequences are indexed by ordinals and not cardinals. The question is not "how many digits are there" but "how long is the sequence of digits". Jun 22, 2019 at 18:30

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).

• I've edited the question to make it a bit clearer. Jun 22, 2019 at 16:55
• @GezaKerecsenyi See my edit. Jun 22, 2019 at 17:13