A sequence that repeats after n terms, as n goes to infinity

Factual Questions
1

I have a sequence that repeats after one term:

1,1,1,1,1,1,1,1,1…

I have a sequence that repeats after two terms:

1,2,1,2,1,2,1,2,1,2…

Three terms:

1,2,3,1,2,3,1,2,3… You get the idea.

A sequence can repeat after n terms for any arbitrarily large but finite n.
But can I claim to have a sequence, hidden behind that curtain over there, which repeats after an infinite number of terms? And not, I should add, by already repeating after a finite number of terms, like the sequence 1,2,1,2,… which repeats after 2 or 4 or 6… or 2n terms, including n = infinity. For my magic sequence there is no repetition until an infinite number of terms.

Is such a thing coherent, or does it break a definition somewhere?

As a followup, if I can by fiat construct a sequence that repeats after an infinite number of terms - if such a thing can exist - how could I know that any given sequence that does not appear to ever repeat (like, say, the natural numbers or the decimal digits of sqrt2) does not, in fact, actually repeat after an infinite number of terms?

2

You can have transfinite sequences, but they are not ordinary sequences. If that is OK then, sure, 1,2,3,…,1,2,3… is a sequence that repeats after infinitely many steps.

Decimal expansions are just normal sequences; how are you defining transfinite extensions of them?

3

I may come to regret asking this, but… what is the difference between a sequence that repeats after infinitely many steps and a sequence that never repeats at all? How can you tell the two apart?

4

In the sense @DPRK is using it, a sequence is indexed by an ordered set that you need to specify. For the sequences you’re used to thinking of, the indexing set is \mathbb{N}, the set of natural numbers (positive integers). But the example DPRK gives, of a sequence that repeats after infinitely many steps, is indexed by the set \mathbb{N}\cup\mathbb{N}, which is a copy of the positive integers followed by another copy.

Decimal expansions are, by definition, indexed by \mathbb{N}. There’s no obvious way of giving meaning to a real number whose “decimal expansion” is given by a sequence indexed on \mathbb{N}\cup\mathbb{N}. Presumably you could define a set of “numbers” given by such sequences (though there’s no immediately obvious way of defining arithmetic), but they won’t be the usual real numbers.

5

Well, I’m out of my depth here, but keeping to a somewhat concrete example, take the square root of 2:

1.41421356etc

Starting from the ‘standard’ square root, I assert that I construct a new number, call it brossaroot 2, which looks like square root 2 , but after an infinite number of decimal places the sequence 141421356… appears again. And again, after another infinite number of decimal places, and on and on ad infinitum. So, in a vague sense, brossaroot2 contains infinite copies of its infinite base ten digits. It’s still not rational, though, because the length of the repeat is infinite.

Can brossaroot2 exist? If so, how do I know that brossaroot2 and regular square root 2 aren’t the same thing?

6

Thank you @brossa that’s what I was trying to get at

7

The sort of thing I’m thinking of isn’t just two copies, though, which is what I assume N∪N means - it’s an infinite number of copies, which I don’t know how to denote. N∪N… perhaps? There’s probably a more elegant expression.

8

The thing is that sequences indexed on \mathbb{N} are simply a way of representing real numbers, and the representation is defined by saying that a sequence 0.a_1a_2a_3\ldots represents the real number \sum_{i=1}^\infty a_i 10^{-i} (along with discussing limits and the meaning of infinite sums). If you want to say how some other kind of sequence represents a real number, it’s incumbent on you to say how it represents a number. There isn’t an obvious way to do that with other types of sequences.

I would write \mathbb{N}^{\mathbb{N}} (or \omega^\omega), but I’m not sure if that’s the common notation.

9

I think the OP is missing the subtlety that DPRK introduced.

In the set of positive integers and decimal expansions “infinite” means without end. Therefore there is no possible “after” to an infinite sequence. It does break the definition. Nothing like brossaroot2 is possible.

DPRK is pointing out that math is much larger than that simple sequence. By using set theory, which, basically, all math is founded on, new and different kinds of sets can be created that would give an answer to the question, although that answer is not something findable with the everyday math non-mathematicians are familiar with.

Explaining math without using math is problematic, I know. But I think totally unfamiliar terminology often adds confusion rather than clears things up.

10

I’m sure that I am missing subtleties! I’m afraid that this is starting to sound like I’m asking what infinity plus one is.

Here’s my proposed method to arrive at brossaroot2:

*Start with the first n digits of square root 2, and repeat them.
n=2 1.4141414…,
n=3 1.41141141…,
n=4 1.41414141…(that’s a sneaky one!)
n=5 1.414214142…

brossaroot2 would be the limit when n=>infinity.

It seems to me that I can construct an approximation to brossaroot2 for any arbitrarily large n, and find the resulting approximation on the line of the reals. What it that makes ‘actual’ brossaroot2 not appear on the number line?

It would be fair to answer ‘because repetition after an infinite number of terms is logically incoherent’

11

If I’m understanding correctly, this would simply converge to the usual \sqrt{2} = 1.41421356\ldots. Each term after n agrees with \sqrt{2} in its first n digits, which implies that the limit is \sqrt{2}. I don’t see how this sequence of numbers gets at what you’re really trying to do.

There are ways to make it logically coherent, as suggested by DPRK and me, but you lose the connection to the real numbers.

12

Or, even more simply, after infinity is meaningless for a decimal expansion.

Remember the many threads about claiming that 0.9999… was not equal to 1? The mistake there was thinking there was a point where the sequence stopped, leaving a gap. You’re also saying that at some point your sequence stops. You’re calling that point infinity but there is no such point. The sequence never stops so there cannot be an after in which to stick things.

Simulposted with Topologist. Yes, the limit of your sequence is the square root of two. However, at any finite n, your sequence is neither the square root of two or anything else. It’s just the square root of two with a possibly very long finite decimal set interpolated. The limit is already “at” infinity so it can’t have another series of numbers appended afterward.

Yes, this is why infinity is handled only mathematically. You can’t get there from there.

Edited to correct a mistake.

13

By the way, Harpo and I are not really disagreeing. He’s saying that, for decimal expansions of the reals, there is no sense in which there is a place to add more digits “after infinity.” I’m saying that there are perfectly good ways to define such sequences, but they no longer have a connection to the real numbers.

14

Yes, I hoped I had made that clear. But brossa is clearly sticking to the reals, even just the rationals.

15

I appreciate your responses - I think I’m closer to understanding what you’re getting at. I’m sure I’ll fully understand it after rereading an infinite number of times.

16

It comes down to what we really mean by a sequence of digits?

When mathematicians talk about sequences of digits what they mean is a mapping that associates every natural number [1,2,…,infinity) (not including infinity itself) with a digit {0, 1, 2, … 9} that represents the value at that point in the sequence.

For example the non-repeating sequence of pi 3, 1, 4, 1, 5, 9,… is represented by the map

1 → 3
2 → 1
3 → 4
4 → 1
5 → 5
etc.

Once pi fills in all of the places from 1 up to a non inclusive infinity, that is it. There is no place to map the infinity plus 1 digit.

Now if you wanted to you could expand the notion of sequence to be a mapping between a transfinite set and the digits from 1 to 10, but that would be a non-standard definition of sequence, and certainly would not be relatable to a decimal expansion of a real number

17

As has been explained, if x is a real number and, e.g., 0<x<1, then x has a decimal expansion

0.a_1a_2a_3\cdots = \sum_{i\in\mathbb{N}}a_i10^{-i}

where \mathbb{N} = \{\,1,\,2,\,3,\,4,\,\ldots\,\} so there are no digits remaining or defined after an infinite number of decimal places. The natural numbers correspond to the first infinite ordinal, often denoted \omega. After that come bigger ordinals like \omega+1, \omega+2, \omega+\omega, \omega^2, \omega^\omega, \omega^{\omega^{\omega^{\cdot^{\cdot^\cdot}}}} but we have left the natural numbers behind, even though these are still countable.

Now to the point, there was a short article by Lightstone in the American Mathematical Monthly, 1972, where he proposed extending the real numbers to hyperreal numbers, as in non-standard analysis, and considering decimal expansions of those. The thing to watch for here is that the hyperreals/hyper natural numbers are not the same as real/natural numbers: there are more of them. In particular, there are infinitely large numbers. A hyperreal between 0 and 1 will have a decimal expansion \sum_{i\in{^*\mathbb{N}}}a_i10^{-i} which he writes 0.a_1a_2a_3\cdots;\cdots a_H \cdots but this is not as nice as you might want, because not every function ^*\mathbb{N} \to \{0,\,1,\,2,\,3,\,\ldots,\,9\} represents the decimal expansion of a hyperreal number, nor is it clear how to break up a given decimal expansion into real and infinitesimal parts. In any case, to reiterate, you are no longer dealing with real numbers.