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I have seen tests used to describe converging sequences as sublinear and logarithmic. Are there tests for sequences that converge even slower, and if so what are they?

For clarity, I am looking at sequences (not series), and only those that converge. For further context, see https://en.wikipedia.org/wiki/Rate_of_convergence

Ret
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1 Answers1

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You can come up with many such tests using the comparison test and the following two sequences of series. The following are all divergent:

$$ \frac{1}{n}, \frac{1}{n\log n}, \frac{1}{n\log n\log\log n}, \frac{1}{n\log n\log\log n\log\log\log n}, \cdots $$

The following are all convergent:

$$ \frac{1}{n^2}, \frac{1}{n\log^2 n}, \frac{1}{n\log n(\log\log n)^2}, \frac{1}{n\log n\log\log n(\log\log\log n)^2}, \cdots $$

Unfortunately, there is no perfect convergence test, that is, there is no one series "in the middle", what was called by du Bois-Reymond a Pantachie (a Greek word). In fact, due to classical results by Hausdorff, we know that there are not even optimal sequences of convergence tests; this is known as a Hausdorff gap. In other words, the sequences described above are not enough to determine the status of all possible series, and there is no way to extend them do ones that do.

So far we have been considering arbitrary series. One might ask whether the situation changes when the series are succinctly described. For example, what happens if we restrict our attention to recursive series, that is ones computed by a program? To avoid some technicalities, we can assume that the entries are all negative (whole) powers of two; every series can be rounded to such a series without affecting convergence or divergence. In that case there trivially is an optimal sequence of tests, simply because the recursive series are countable. However, if we require that the sequence of tests be generated by a single program, then again there is no optimal sequence of tests, and this can be proved using the constructions in the unrestricted case.

The next step is to look for more restricted classes of series for which the program is tractable. For example, what happens if we consider logarithmico-exponential series? Unfortunately I don't know the answer.

Coming back to the unrestricted case, we mentioned that there is no optimal sequence of tests. But there could be an optimal well-ordered scale consisting, say, of series diverging ever more slowly, eventually beating all divergent series. Surprisingly, the existence of scales is independent of ZFC. This is related to the so-called Cichoń's diagram.

Summarizing, this innocent question opens up a window into several mathematical areas: set theory, recursion theory, and even analysis.

Yuval Filmus
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  • While I appreciate your response, I am specifically looking at sequences, not series. If the answer to my question is wrapped up in your answer, I'm afraid I don't understand the connection. – Ret Aug 09 '14 at 03:17
  • Yes, the questions are closely related. The case usually discussed in the literature, and more or less equivalent to convergent series, is sequences tending to infinity. Given a convergent sequence, subtract the limit , remove the signs and take reciprocals to obtain a sequence tending to infinity. Round it to integers and you're in the classical setting. The upshot is that there is no sequence converging as slow as possible, neither is there such an optimal sequence of sequences covering ever slower; but an uncountable scale might exist. – Yuval Filmus Aug 09 '14 at 03:44
  • So going from your answer, I take the sequences of the partial sums of the divergent series, then invert the elements. Since the series you gave diverge slower and slower, the sequences I will get converge slower and slower. But since there is no "slowest divergent series", I won't find a "slowest convergent sequence". Furthermore, for the test I was looking for, I can compare the sequences (by way of the inverse of the sequence of partial sums) to the ever-slowing series. Have I understood this correctly? If so, I thank you very much. – Ret Aug 10 '14 at 04:13
  • Yes, I believe you understood me correctly. – Yuval Filmus Aug 15 '14 at 17:34