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The following is a FAQ that I sometimes get asked, and it occurred to me that I do not have an answer that I am completely satisfied with. In Rudin's Principles of Mathematical Analysis, following Theorem 3.29, he writes:

One might thus be led to conjecture that there is a limiting situation of some sort, a “boundary” with all convergent series on one side, all divergent series on the other side—at least as far as series with monotonic coefficients are concerned. This notion of “boundary” is of course quite vague. The point we wish to make is this: No matter how we make this notion precise, the conjecture is false. Exercises 11(b) and 12(b) may serve as illustrations.

Exercise 11(b) states that if $\sum_n a_n$ is a divergent series of positive reals, then $\sum_n a_n/s_n$ also diverges, where $s_n = \sum_{i=1}^n a_n$. Exercise 12(b) states that if $\sum_n a_n$ is a convergent series of positive reals, then $\sum_n a_n/\sqrt{r_n}$ converges, where $r_n = \sum_{i\ge n} a_i$.

Although these two exercises are suggestive, they are not enough to convince me of Rudin’s strong claim that no matter how we make this notion precise, the conjecture is false. Are there any stronger theorems in this direction?

Edit. For example, are there any theorems about the topology/geometry of the spaces of all convergent/divergent series, where a series is viewed as a point in $\mathbb{R}^\infty$ or $(\mathbb{R}^+)^\infty$ in the obvious way?

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    $\begingroup$ Nice question; that claim has always bothered me. $\endgroup$ Commented Dec 14, 2010 at 19:05
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    $\begingroup$ Perhaps a partial answer could be found within the theory of Abel's summations and Tauberian theorem. But my knowledge of these topics is not accurate enough. $\endgroup$ Commented Dec 15, 2010 at 7:34
  • $\begingroup$ We could define the topology to consist of the set of all sequences, the empty set, the set of such that the associated series converges, and the series where the set the associated series diverge. With this topology, their is a well defined boundary which is the empty set. Not that this helps. $\endgroup$ Commented Jun 4, 2013 at 4:25
  • $\begingroup$ home.mathematik.uni-freiburg.de/mildenberger/postings/paperspdf/… $\endgroup$ Commented Sep 27, 2022 at 13:09

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A rather detailed discussion of the subject can be found in Knopp's Theory and Application of Infinite Series (see § 41, pp. 298–305). He mentions that the idea of a possible boundary between convergent and divergent series was suggested by du Bois-Reymond. There are many negative (and mostly elementary) results showing that no such boundary, in whatever sense it might be defined, can exist.

Stieltjes observed that for an arbitrary monotone decreasing sequence $(\epsilon_n)$ with the limit $0$, there exist a convergent series $\sum c_n$ and a divergent series $\sum d_n$ such that $c_n=\epsilon_nd_n$. (This can be easily deduced from the Abel–Dini theorem).

Pringsheim remarked that, for a convergent and a divergent series with positive terms, the ratio $c_n/d_n$ can assume all possible values, since one may have simultaneously $$\liminf\frac{c_n}{d_n}=0\qquad\mbox {and}\qquad\limsup\frac{c_n}{d_n}=\infty.$$

I like the following geometric interpretation. Given a (convergent or divergent) series $\sum a_n$, let's mark the sequence of points $(n,a_n)\in\mathbb R^2$ and join the consecutive points by straight segments. Then there is a convergent series $\sum c_n$ and a divergent series $\sum d_n$ (both with positive and monotonically decreasing terms) such that the corresponding polygonal graphs can intersect in an indefinite number of points.

The results remain essentially unaltered even if one requires that both sequences $(c_n)$ and $(d_n)$ are fully monotone, which is a very strong monotonicity assumption. This was shown by Hahn ("Über Reihen mit monoton abnehmenden Gliedern", Monatsh. für Math., Vol. 33 (1923), pp. 121–134).

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    $\begingroup$ Thanks. Rudin actually mentions Knopp, but for some reason I always thought Rudin was giving Knopp as a general reference and not as a specific reference for this particular question. I should have at least checked out the library's copy of Knopp first! $\endgroup$ Commented Dec 14, 2010 at 21:37
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    $\begingroup$ You are welcome. The book is a gem. $\endgroup$ Commented Dec 14, 2010 at 21:46
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The following exercise appears in Bruce Driver's analysis lecture notes (Exercise 25.23, page 513).

Proposition. There does not exist a sequence $\{a_n\}$ such that, for all sequences $\{\lambda_n\}$, $\sum_n |\lambda_n| < \infty$ iff $\sup_n |a_n^{-1} \lambda_n| < \infty$.

The proof goes by showing that $\{\lambda_n \} \mapsto \{a_n \lambda_n\}$ would give a bijective bounded linear operator from $\ell^\infty$ to $\ell^1$. By the open mapping theorem this would be a homeomorphism, which is absurd.

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    $\begingroup$ This can exercise can also be found in Folland's Real Analysis book in the section on the Baire category theorem. Great question and great answer! $\endgroup$ Commented Dec 14, 2010 at 18:53
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    $\begingroup$ Suppose there is such a sequence ${a_n}$. It follows that $\sum_n |a_n|<\infty$, contradicting the above mentioned exercise 12(b). So in what sense is this an answer? $\endgroup$
    – Guntram
    Commented Dec 14, 2010 at 19:22
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    $\begingroup$ @Guntram: Hmm, good point. I read this proposition as "there is no slowest rate of decay for summable sequences", but you are quite right that it follows from 12(b), which I had not noticed. SO it is not in fact stronger. $\endgroup$ Commented Dec 14, 2010 at 19:49
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An amusing exercise I like to set from time to time is this. As is well known, the sum of $1/n$ diverges, as does the sum of $1/n\log n$ and the sum of $1/n\log n\log\log n$, and so on. But what happens if we define $f_k(n)$ to be $1/n\log n\dots\log_kn$, where $\log_kn$ stands for the k-fold iterated logarithm, and $k$ is maximal such that $f_k(n)\geq 1$? I'm not asking for the answer -- just drawing attention to this function that's close to the non-existent boundary. (A follow-up question might be to find a reasonably natural function that is even closer. For instance, can one define $f_\alpha(n)$ for some very large countable ordinals $\alpha$?)

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    $\begingroup$ I would imagine that a natural way of making rigorous the notion of "boundary" would be by introducing some kind of (ordinal) rank on series, and then a boundary would be a natural gap in the resulting pre-ordering. $\endgroup$ Commented Dec 14, 2010 at 22:16
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    $\begingroup$ I am a little confused by the statement of this excercise. Are we considering a single function $f(n):=1/n\log{n}\ldots\log_k{n}$, where $k$ is maximal such that $\log_k{n}\geq 1$? This question (in a slightly different form) appeared as A4 on the 2008 Putnam. $\endgroup$ Commented Dec 15, 2010 at 7:49
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    $\begingroup$ This was also Problem E3381 in the American Mathematical Monthly. The solution (AMM 1992, page 165-166) is followed by an editorial note mentioning a paper by Reingold and Shen (More nearly optimal algorithms for unbounded searching. II. The transfinite case. MR1082143)) discussing a hierarchy of slowly converging series of height $\epsilon_0$, apparently with motiviation from theoretical Computer Science. $\endgroup$
    – Goldstern
    Commented May 7, 2011 at 17:33
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As Dylan Wilson pointed out, the following question appears in Folland's real analysis book:

(second edition, pg. 164) (33) There is no slowest rate of decay of the terms of an absolutely convergent sequence; that is, there is no sequence $\{ a_n \}$ of positive numbers such that $\sum a_n|c_n| < \infty$ iff $\{ c_n \}$ is bounded.

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  • $\begingroup$ I find this very satisfying, implying that the gap between convergence and divergence is not really so subtle as it might at first seem: it is actually infinitely wide in some sense, although not calculably so (reminding me of the Halting Problem, which is also not calculable). I wonder whether this "slowest rate of decay" concept can also be applied in some way to the boundary of the Mandelbrot set and other such infinitely and not trivially self-similar shapes: that is, that it is not possible to calculate what such a set will look like except to approximately construct it. $\endgroup$ Commented Nov 11, 2019 at 21:38

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