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Inspired by Are half of all numbers odd?, idea of asymptotic density in this answer for odd numbers in set of all positive integers as $$\lim_{n \to \infty} \frac{|\{k \leq n, k \text{ is odd}\}|}{n},$$ (which is $1/2$), which can be viewed as a probability that random integer is odd (if we limit to numbers up to finite boundary $n$ and then going $n\to\infty$). In same spirit, I was wondering what is a probability that random sequence $\{a_k\}_{k=1}^\infty$ of real numbers is convergent? How do we even choose probability distribution (as in above case, it should be in some sense uniform)? Intuitively, I would expect the probability to be $0$ since there seems to be vastly more divergent sequences than convergent, but I was thinking the same about transcendental numbers, so... Also, not sure if this will be any helpful, but cardinality of convergent sequences is continuum, as this shows: What is the cardinality of the set of all sequences in $\mathbb{R}$ converging to a real number? ...

I wanted to try similar approach as above with odd integers, to that I would need to have some sequence of finite sets of sequences $A_k$ and $B_k$, where $A_k$ converges to set of all convergent sequences and $B_k$ converges to set of all sequences (and I am not even sure in what metric space that would be...), and then compute $\lim \frac{|A_k|}{|B_k|}$. Problem is that I fail to construct sequence that would satisfy even the finiteness property, not mentioning the convergence...

Is there a natural way to define probability of random sequence being convergent? And if so, what is it?

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    $\begingroup$ Surely the answer, if any answer could be defined, would be $0$. Why should a random sequence even be Cauchy? If your definition allows for a positive probability that $|a_1-a_0|>\epsilon$, and if you assume the same distribution at each slot independently, then the probability that a random sequence is Cauchy is $0$ (assuming it can be defined). Or did I misunderstand your goal? $\endgroup$ – lulu Sep 23 '18 at 19:53
  • $\begingroup$ @lulu I didn't think of it in probabilities, but it seems good way to think about it (also the density of odd integers actually makes more sense now to me when I ask what is probability that random number is odd...). I guess you could put that as an answer if you like. $\endgroup$ – Sil Sep 23 '18 at 19:57
  • $\begingroup$ Not precise enough for an answer, I think. Yes, probabilities are a good approach., I'd say. Your even/odd case works that way. It's sensible to imagine that "probability" there has to mean "limit the choice to $N$ or less, and then let $N\to \infty$. Here, though, I can't think of anything that would give a non-trivial result. $\endgroup$ – lulu Sep 23 '18 at 20:03
  • $\begingroup$ @lulu I've restated the question in terms of probabilities, but I see your point, probably anything else than $0$ cannot be expected... So it just boils down into how to define it properly... $\endgroup$ – Sil Sep 23 '18 at 20:19

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