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This question might seem too fuzzy, and if so, I will be happy to withdraw it. Until then, here it is:

I know that a method of slowing a divergent series of positive reals is to replace the $n$-th term by it divided by the first $n$ terms. In this way the series obtained stays divergent, but it decreases infinitely faster.

Now consider the class $\Sigma$ of divergent series made of positive reals, where $(a_i)<(b_i)$ means that $\lim_{i\to\infty}a_i/b_i=0$. Consider now a decreasing sequence of series.

The question is if there is a notion of convergence fit to this order (most plausibly in a weak/generalized sense), and if there is an extension of $\Sigma$ where one could give some meaning to "the slowest divergent series".

I suspect that the/some answer would have to do with something like nonstandard analysis: one might then reframe even the definition of the order relation, in the natural way.. I would highly appreciate other speculations about the statement of the problem too.

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I know that there is no slowest rate of decay for convergent series, in the sense that there is no sequence $a_n$ of positive real numbers such that $\sum_n a_n |b_n| < \infty$ if and only if ${b_n}$ is a bounded sequence. This seems related to your question, but I don't think it answers it. –  Paul Siegel Dec 15 '11 at 11:54
    
Let $a_n$ be a convergent series. We can consider series of the form $b_n = a_{f(n)}/g(f(n))$ where $g(n)$ is a sequence that increases really fast and $f(n)$ increases correspondingly slowly. If we choose $f$ and $g$ correctly then $\sum_{i=1}^n b_n \leq \sum_{i=1}^{f(n)} a_n$ for all $a_n$. A minimal element would have to be stable under all such transforms of this type. –  Will Sawin Dec 24 '11 at 3:25
    
More simply, let $s_n$ decrease to $0$, then there is always an $a_n$ so that $\sum a_ns_n$ is convergent but $\sum a_n$ is divergent. It seems unlikely that you could therefore have any good notion of minimality. –  Will Sawin Dec 24 '11 at 3:30

2 Answers 2

This question was considered by du Bois-Reymond in 1870 and he came to some conclusions about convergence classes being linearly ordered that were not well substantiated. Hausdorff then considered the question of pantachies (maximal linearly ordered convergence classes) in great detail around 1908 and cleared up some of the ambiguities left by du Bois-Reymond. Hardy also considered this question from a different perspective. In modern terms the cardinal invariants b and d play a crtitcal role.

However, the upshot is, I believe, that no notion of least divergence class leads to any interesting concept. Non standard analysis will not help much here. For example, non-standard models obtained as ultrapowers will linearly order the convergence classes but their cofinality is non-trivial and leads to an interesting field of study.

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Dear Juris, thank you for the references! It seems interesting to know more about Hausdorff's paper, and about what b,d are. –  Mircea Dec 15 '11 at 12:08
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To learn about b and d you could look at Blass' article in the Handbook of Set Theory. You can find references on du Bois-Reymond and Hausdorff in my article on the History of the Continuum in Volume 6 of the History of Logic. –  Juris Steprans Dec 15 '11 at 13:40
    
Thanks for this. I will come back as soon as I read enough of it. –  Mircea Dec 15 '11 at 15:42

Let $a_n$ and $b_n$ be two strictly increasing sequences of natural numbers,with $\sum_{n=1}^{\infty}\frac{1}{a_n}=\infty$ and $\sum_{n=1}^{\infty}\frac{1}{b_n}=\infty$

.When I was an undergraduate student, i had in mind the ''conjecture'' that using the above sequences it is impossible to have $\sum_{n=1}^{\infty}\frac{1}{a_n+b_n}=c, c\in R$.

Surprisingly enough, i believe we can find a counterexample: We can find two sequences $a_n,b_n$ with $\sum_{n=1}^{\infty}\frac{1}{a_n}=\infty,$ $\sum_{n=1}^{\infty}\frac{1}{b_n}=\infty$ but $\sum_{n=1}^{\infty}\frac{1}{a_n+b_n}=1$ (or some other constant ,but this is a specific example)

Let $a_n+b_n=2^n$

The idea is to increase the sum of the first series by a number aproximately $ln2$ between the $2^{2^{.^{.{^2}}}}$

($n-2$ exponents) and $2^{2^{.^{.{^2}}}}$ ($n$ exponents) term. The same idea will also run for the second series.

$a_1=b_1=1\\a_2=b_2=2\\a_3=5,b_3=3\\a_4=12,b_4=4$

because $a_4$ is ''too big'' we will increase $a_n$ only by $1$ for the next $12$ terms,and increase significally $b_n$ in the following way:

$a_5=13 , b_5=19,a_6=14, b_6=50 ...$ and so on ,until we reach $ a_{16}=24 ,b_{16}=65512 $$.$ Again we will switch the growth of $b_n$ and $a_n,$ adding only $1$ to $b_n$ for the next $65512$ terms but increase significally $a_n$ in order to sum $2^n$. So , we will reach at $2^{2^{2^{2}}}=2^{65536}=a_{65536}+b_{65536},$ with $a_{65536}=2^{65536}-131024$ and $b_{65536}=131024$ and then continue in the same way.

It easy to see that when we reach $2^{2^{.^{.{^2}}}}$ ($2n$ exponents) both $\sum{\frac{1}{a_n}}$ and $\sum{\frac{1}{b_n}}$ will be at least $nln2$ and so, conclude that both series reach infinity.

We could define a set $A$ of pairs of $(\sum{\frac{1}{a_n}}$,$\sum{\frac{1}{b_n}})$ both being divergent, but with the property that $\sum_{n=1}^{\infty}\frac{1}{a_n+b_n}$ is convergent. I believe that this will define the ''pair of slowest divergent series''

Note:we use $''ln2''$ because $\frac{1}{n+1}+...+\frac{1}{2n}$ has limit $ln2$

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protected by François G. Dorais Oct 7 '13 at 1:27

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