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In what sense is the sequence $(0,1,2,0,1,2,\ldots)$ convergent?

A problem I'm working on needs a concept of convergence which understands the fact that this sequence extended infinitely is convergent - at least in the sense that it is stable and orbits a fixed point.

The name may not be "convergence".

I can of course engineer something but I presume there is some accepted terminology I can use.

For example I want to describe that:

$$\lim_{n\to\infty}\left((1-2^{-n})+ (n\mod3)\right)$$ converges upon an orbit centred on the number $2$

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    $\begingroup$ There are three convergent subsequences. One converges to 0, the other converges to 1 and… $\endgroup$ May 6 '17 at 16:13
  • $\begingroup$ Thanks @LiChunMin that's exactly the answer I need. The problem's a little more complicated in that there are infinitely many subsequences... every one of which converges, but since there are infinitely many, there are new subsequences constantly appearing. $\endgroup$ May 6 '17 at 16:34
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    $\begingroup$ Aha. Sure there are infinitely many convergent subsequence there. For each n, make put n zeros there and make the tail to be all one… That lead me to think of this question: are there only countably many convergent subsequences? $\endgroup$ May 6 '17 at 16:44
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    $\begingroup$ @LiChunMin In a sequential space like $\mathbb R$, the set of limits of subsequences is the same as the set of "accumulation points" of the sequence. See my answer for some relevant links. $\endgroup$
    – Mark S.
    May 6 '17 at 17:58
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The set of "accumulation points" (sometimes "cluster points") of the first sequence is $\{0,1,2\}$ and that set for the second sequence is $\{1,2,3\}$. This is because every neighborhood of one of the numbers, say $1$, has infinitely many sequence entries in that neighborhood.

This idea of the accumulation points is closely related to other ideas of "limit set" like attractors and limit cycles.

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I would not use the word "convergent" in any sense! This sequence is "periodic" with period 3.

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  • $\begingroup$ The second sequence is not periodic, yet it has an important relationship to convergence since there is a finite set of limits of convergent subsequences. $\endgroup$
    – Mark S.
    May 6 '17 at 18:00
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You can take the already defined Cesàro summability, and sort of modify it to fit your needs. Let $(a_n)_{n=1}^\infty$ be your sequence. We can define a sequence to be Cesàro convergent if and only if

$$ \lim_{n\to \infty} \frac{1}{n}\sum_{i=1}^n a_i \in \mathbb R$$ That is, if the limit above exists, the sequence is Cesàro convergent.

Notice, in your case, the limit above converges to $1$, since that is the average value of your sequence. So, in this sense, the sequence is "Cesàro convergent around 1".

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