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.

  • 1
    $\begingroup$ 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. $\endgroup$ Dec 15, 2011 at 11:54
  • $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Dec 24, 2011 at 3:25
  • $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Dec 24, 2011 at 3:30

3 Answers 3


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.

  • $\begingroup$ Dear Juris, thank you for the references! It seems interesting to know more about Hausdorff's paper, and about what b,d are. $\endgroup$
    – Mircea
    Dec 15, 2011 at 12:08
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    $\begingroup$ 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. $\endgroup$ Dec 15, 2011 at 13:40
  • $\begingroup$ Thanks for this. I will come back as soon as I read enough of it. $\endgroup$
    – Mircea
    Dec 15, 2011 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.


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 $ln2$


Here is an interesting candidate for the slowest diverging family of sequences. It is well known that every $a_n$ in the family





has the property that they converge when summed reciprocally, so to have any chance of being the fastest it would need to be able to beat all of these. In this discussion, we will assume that all logarithms are taken base $\delta$ (for some fixed delta) which we will come back to choosing later. We will also use the notation


so that $\log^+(\log^+(\log^+(...(\log^+(x)))))$ is always defined and real. We also let $\log^{(m)}(x)$ denote the $m-$fold application of the $\log^{+}$ function on $x$. We now define

$$\mathscr{L}_k(x):=n\left(\prod_{n\leq k}\log^{(n)}(x)\right)\log^{(k)}(x)$$

so that $\mathscr{L}_0(x)=n$, $\mathscr{L}_1(x)=n\log^2(n)$, etc... We now defined the very natural "closure of these" functions


note that we dropped the extra $\log^{(k)}(x)$ factor when taking the limit, since it will always approach $1$, so


for any choice $k$. The proof of the above statement is more complicated than the handwaving in the present answer, but it is not very hard at all. Before showing that


converges we need some preliminaries. We begin by noticing that for $x>1$

\begin{align*} \mathscr{L}_{\infty}\left(\delta^x\right)&=\delta^x\left(\prod_{n=1}^{\infty}\log^{(n)}\left(\delta^x\right)\right)\\ &=\delta^x x\left(\prod_{n=1}^{\infty}\log^{(n)}\left(x\right)\right)\\ &=\delta^x\mathscr{L}_{\infty}(x)\tag{1} \end{align*}

where we used that for $m>1$ $\log^{(m)}\left(\delta^x\right)=\log^{(m-1)}\left(x\right)$ uniformly, and $\log^{+}\left(\delta^x\right)=x$ since $x>1$. We now move on to proving the following theorem:

The series \begin{equation} \sum_{n}\frac{1}{\mathscr{L}_{\infty}\left(n\right)}\tag{2} \end{equation} converges $\mathbf{if\,\,and\,\,only\,\,if}$ $\delta<e$.

To do this, we use the integral test to see that the series in (2) converges if and only if the integral


converges as $x\to\infty$. Using basic calculus tricks we see that for $x>\delta$

\begin{align*} I\left(x\right)&=\int_{\delta}^{x}\frac{1}{\mathscr{L}_{\infty}\left(t\right)}dt\\ &=\int_{1}^{\log(x)}\frac{\ln(\delta)\delta^t}{\mathscr{L}_{\infty}\left(\delta^{t}\right)}dt\\ &=\int_{1}^{\log(x)}\frac{\ln(\delta)}{\mathscr{L}_{\infty}\left(t\right)}dt\\ &=\int_{1}^{\delta}\frac{\ln(\delta)}{\mathscr{L}_{\infty}\left(t\right)}dt+\int_{\delta}^{\log(x)}\frac{\ln(\delta)}{\mathscr{L}_{\infty}\left(t\right)}dt\\ \end{align*}

for $1<x<\delta$, $\log(x)<1$ and so $\mathscr{L}_{\infty}\left(t\right)=t$, meaning that


if $\log^{(m)}(x)\geq1$ but $\log^{(m+1)}(x)<1$, then we can apply the above formula $m$ times to get that

\begin{align*} I(x)&=\sum_{j=1}^m\ln^{j+1}(\delta)+\ln^m(\delta)I\left(\log^{(m)}(x)\right)\\ &=\sum_{j=1}^m\ln^{j+1}(\delta)+O\left(\ln^m(\delta)\right) \end{align*}

thus, as $x\to\infty$, $I(x)$ will converge if and only if the series


converges which will take place only when $\ln(\delta)<1\implies \delta<e$. This completes our proof.

One interesting thing to notice is that for any of the finite $\mathscr{L}_{k}\left(x\right)$ the base does not matter since they will all be off by at most a constant factor. A reason this family of functions $\mathscr{L}_{\infty}\left(x\right)$ is such a strong candidate is that we have endpoint failure at $\delta\geq e$, which shows that we have a very small space to work with when improving.


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