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For any real number $x$ we can construct the sequence $a_n = \lfloor n \cdot x \rfloor - \lfloor (n - 1) \cdot x \rfloor$ for $n \geq 1$ and we have that the value of $a_n$ is either $\lfloor x \rfloor$ or $\lceil x \rceil$ and $$ \lim_{n \to \infty} \frac1n \sum_{i=1}^n a_i = \lim_{n \to \infty} \frac{\lfloor n \cdot x \rfloor}{n} = x $$ so we have a bounded sequence whose average sequence converges to $x$. Now my question: is there any way to know that the average sequence of $a_n$ converges by looking directly at $a_n$ and not dealing with the averages directly? In other words I am looking for sufficient (and necessary, if possible) conditions for this convergence.

For context, I encountered this problem as part of my attempt to construct the real numbers as averages of infinite sequences of integers. This is inspired by [1], where the reals are constructed as slopes of almost linear integer functions. The blocking point for my case is that, given $a, b \in \mathbb R$ and sequences $a_n, b_n \in \mathbb Z$ such that their average sequences converge to $a$ and $b$, respectively, I need to either:

  • define the representation of the product $a \cdot b$ as $c_n = a_n \cdot b_n$ and manage to prove that its average sequence converges, or
  • find a different way to define the representation of the product, which would allow a convergence proof.

[1] https://arxiv.org/abs/math/0301015

Cristian Gratie
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  • Can you find a sequence of 0 and 1 whose average does not converge? That might give you some idea as to the intricacy of the question that you're asking. – Calvin Lin Mar 11 '22 at 23:46
  • Yes, I have already seen this question, but could not extract a useful intuition from that yet. One thing I tried (inspired from the paper I mentioned) is to translate the condition of almost linear function to my case. This gives that $\sum_{k=1}^n a_{i+k} - \sum_{k=1}^n a_{j+k}$ should be bounded. My intuition is that this should ensure convergence of the average sequence (though I don't have a proof yet) but unfortunately this condition is not preserved by element-wise products. – Cristian Gratie Mar 11 '22 at 23:59

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