## Defining the slowest divergent series

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 2011 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 2011 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 2011 at 3:30