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It is well known that there is no slowest converging infinite series (see e.g. here).

But there is also no largest rational number whose square <=2. Once we complete the rationals to the reals, such a number exists.

Looking at the standard examples of series that diverge ever more slowly: ${1\over n}, {1 \over {n \ log(n)}}, {1\over{n \ log(n) \ log(log(n))}} ... $ you get the feeling that these are tending towards something, even if this something is not itself a series, or it is not a limit in the usual sense.

Is there some notion of completion of the space of infinite sequences, so that some member of the larger space lies exactly at the border of convergence and and divergence?

Edit: One possible way to do this would be via nonstandard analysis. Hyperreals are equivalence classes of sequences, and are totally ordered - i.e. they provide a way to say whether any sequence is larger than any another. So one might expect to find a "boundary" between sequences whose sum converges, and sequences whose sum does not converge.

Specifically, you would totally order non-negative series by comparing the hyperreals defined by their partial sums $A_n = \sum_{m=1}^n a_m$. This order would respect convergence, in the sense that a divergent series could not be "less than" a convergent series, and would also respect convergence speeds, in the sense that if $a_n$ would be less than $b_n$ if $A_n/B_n \to 0$.

However the hyperreals are not complete, so there need be no supremum to the set of sequences whose sum converges. It is possible to complete them, but the question then becomes what sort of objects are these completed hyperreals, and can we gain any intuition from them about our original question concerning convergent and divergent series.

I found a related previous question. But no complete answer.

Neuromath
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  • Speaking only morally, a slowest converging series (even saying modulo multiplication of the terms by something approaching a real constant) seems like it would kind of be analogous to a smallest positive number. Outside of boring cases like the integers, making that concept even a little bit natural requires a very odd/esoteric construction (much less standard than those in nonstandard analysis), so I'm pessimistic about any nice concept for this question. – Mark S. Aug 25 '18 at 14:06
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    See here. – plop Jun 10 '21 at 16:41

1 Answers1

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No.

The question's broad enough that I can't prove the answer's no mathematically; one can't expect to prove something that asks about the existence of some "notion" with certain properties. But (i) there is no standard such notion in current standard mathematics (that's more an empirical fact than a mathematical one), (ii) the space of convergent sequences comes with a natural metric, which is already complete.

Edit: It's been conjectured that maybe non-standard analysis furnishes us with the required notion somehow. I know nothing about nsa, but I doubt it. Hard to see how nsa is going to make the following disappear somehow:

Triviality: If $a_n>0$ and $a_n\to0$ then (i) there exists $b_n>0$ with $b_n\to0$ and $a_n/b_n\to0$ (ii) there exists $b_n>0$ with $b_n\to0$ and $a_n/b_n\to\infty$.

Proof. (i) $b_n=\sqrt{a_n}$, (ii) $b_n=a_n^2$.

David C. Ullrich
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  • Thanks!

    But I wonder if there could be something from nonstandard analysis. We can turn the set of all sequences into a totally ordered set, the hyperreals, after choosing an arbitrary non-principal ultrafilter. The hyperreals are not complete, so the set of all convergent sequences need not have a supremum (although it is conceivable this specific set does have one). But there may be some way to complete this space?

     – Neuromath Aug 22 '18 at 12:42
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    @Neuromath while I don't know much about this, I'd imagine the transfer principle could make applying nonstandard analysis here as a way to "avoid" a problem that exists classically difficult. – Mark Schultz-Wu Aug 23 '18 at 14:40