I have been explaining the Riemann Rearrangement Theorem to a friend of mine, and they feel as though the definition of series using partial sums "doesn't work" for conditionally convergent sequences. I understand how they feel: the RRT feels counter-intuitive, and so the result should be denied as a contradiction and the partial-sum definition rejected.

However, the definition of a series as

$$\sum_{k=1}^{\infty} a_k = \lim_{n\to\infty} \left(\sum_{k=1}^n a_k\right)$$

feels like the most natural approach to take. If I wanted to add infinitely many numbers together by hand, this is how I would have to do it.

Is there an alternative (but not equivalent) definition of an infinite sum that agrees with the partial-sum definition on absolutely convergent sequences, but where the RRT doesn't hold?

My guess is that if you define some value for an infinite series agreeing on absolutely convergent sequences then this method of definition must imply the RRT, but I don't see how that proof could go.

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    $\begingroup$ Since $\mathcal{P}(\mathbb{N})$ and $\mathcal{P}_{fin}(\mathbb{N})$ have different cardinalities ($2^{\aleph_0}$ and $\aleph_0$), I believe there is no way to avoid the RRT-lack-of-commutativity phenomenon when defining series (i.e. infinite sums) through finite sums, no matter if in the standard way, through smoothed/Cesàro sums or whatever. $\endgroup$ – Jack D'Aurizio Oct 28 '18 at 17:43
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    $\begingroup$ I think you're pretty much stuck. There are a huge number of really general notions of summability (e.g. matrix summability methods), and for quite a few of these not only do you still run into these problems, but in fact for pretty much any of these summability methods there are many series for which most rearrangements (in the Baire category sense for certain metrics) of that series have very bad convergence properties (such as every extended real number being a subsequence limit). Or something to this effect . . . $\endgroup$ – Dave L. Renfro Oct 28 '18 at 18:03
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    $\begingroup$ You need to explain to your friend sometimes things make us unhappy and there is not much to do about them. $\endgroup$ – Pedro Tamaroff Oct 28 '18 at 18:41
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    $\begingroup$ It seems the only way to avoid RRT is to declare that only absolutely convergent series may be considered to converge. If a series is only conditionally convergent, you insist that it cannot converge to any value. Basically, by refusing to accept the existence of conditional convergence, you avoid having to deal with its consequences. You also lose any benefit you might have gotten from it. But I think you would still have a consistent system. $\endgroup$ – David K Oct 28 '18 at 23:59
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    $\begingroup$ There is counselling for your friend on most campuses ... $\endgroup$ – zhw. Oct 29 '18 at 0:19

Assume that we want a notion of summability $\sum'$ such that

  1. (Compatibility with absolutely convergent series) $$\sum_{n\geq 1}|a_n|<+\infty\quad\Longrightarrow\quad\sum_{n\geq 1}'|a_n|=\sum_{n\geq 1}|a_n|$$
  2. (Compatibility with the vector space operations and other reasonable assumptions) Provided that both $\sum'a_n$ and $\sum' b_n$ are finite, $\sum'(a_n+\lambda b_n)=\sum' a_n+\lambda\sum' b_n$. Additionally $$ \left|\sum' a_n\right|<+\infty\quad\Longrightarrow\quad \lim_{n\to +\infty}a_n=0,$$ $$ a_n>0,\left|\sum' a_n\right|<+\infty\quad\Longrightarrow\quad \sum' a_n>0$$
  3. (Negation of Riemann-Dini) Provided that $\sum_{n\geq 0}'a_n$ is finite, for any bijective $\sigma:\mathbb{N}\to\mathbb{N}$ $$ \sum_{n\geq 0}'a_n = \sum_{n\geq 0}'a_{\sigma(n)}$$

Then such notion of summability is precisely the notion of absolute-summability. Assume that $\sum'a_n$ is finite and $\{a_n\}$ has an infinite number of both positive and negative terms. By $3.$ we may assume without loss of generality that the sign of $a_n$ agrees with the parity of $n$. If $\sum a_n$ is not absolutely convergent then $\left|\sum a_{2n}\right|$ or $\left|\sum a_{2n+1}\right|$ is unbounded (or both are). By 2. and 3. both $\sum' a_{2n}$ and $\sum' a_{2n+1}$ have to be finite. Assuming that $|\sum a_{2n}|$ is unbounded, by 2. and 3. again $$ \sum_{n\geq 1}'|a_{2n}|\geq \sum_{n=1}^{N}|a_{2n}| $$ has to hold for any $N\in\mathbb{N}^+$, but that implies $\left|\sum' a_{2n}\right|=+\infty.$

  • $\begingroup$ This is exactly what I was looking for/hoping for. Thank you! $\endgroup$ – Santana Afton Oct 29 '18 at 2:32
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    $\begingroup$ The result you are showing is wrong. Consider the summation notion where $\sum'a_n=\sum(a_n-l)$ for any sequence $a_n$ with a limit $l$ such that the latter sum converges absolutely. Then all three of your assumptions are satisfied, but more series are summable. $\endgroup$ – Wojowu Oct 29 '18 at 8:24
  • $\begingroup$ To point out some issues with the argument: you only consider series $a_n$ with infinitely many terms of both signs (my example shows you need to check more). Second, I think the deduction that $\sum'a_{2n}$ is finite is unsound - nothing in the assumptions related convergence of a series and a subseries. I also think there is something fishy about the deduction of the last displayed equation. $\endgroup$ – Wojowu Oct 29 '18 at 8:29
  • $\begingroup$ Note that even if you do only include sequences convergent to 0, you have to somehow treat positive sequences, e.g. you have to exclude the harmonic series from having finite sum $\endgroup$ – Wojowu Oct 29 '18 at 9:02
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    $\begingroup$ @JackD'Aurizio Why does a positive sequence have to have sum greater than the partial sums? Also, I'm you still didn't justify why $\sum'a_{2n}$ is finite. $\endgroup$ – Wojowu Oct 29 '18 at 12:20

In the following definition, the value of the sum clearly does not depend on an ordering and this definition seems at least to some extent reasonable. (And also coincides with the usual convergence for absolute convergent real sequences.)

Definition. Let $(x_n)_{n\in\mathbb N}$ be a sequence of real numbers. We say that $S$ is the sum of the series $x_n$, denoted by $$\sum_{n\in\mathbb N} x_n = S,$$ if and only if for every $\varepsilon>0$ there exists a finite set $F_0\subseteq N$ such that for all finite sets $F\supseteq F_0$ we have $\left| \sum\limits_{n\in F} x_n - S \right| < \varepsilon$. $$(\forall \varepsilon>0) (\exists F_0\text{ finite }) \left(F\text{ is finite and }F\supseteq F_0 \Rightarrow \left| \sum\limits_{n\in F} x_n - S \right| < \varepsilon \right)$$

It is worth mentioning that if we work in a more general setting with sequences in Banach spaces, then equivalence between this definition and absolute convergence no longer holds. (Unconditional convergence and absolute convergence are equivalent for sequences of real numbers, but things can become more complicated if we work with infinite-dimensional spaces.)

This is a special case of a more general definition which is often used if we need summation over arbitrary index set (not necessarily countable). This notion is also mentioned in the Wikipedia article on series in the section Summations over arbitrary index sets (current revision). There are a few posts on this site which contain some basic information about this notion and links to further references. I will simply link to my previous question related to this topic - since in this question I have tried to collect the posts on this site related to this notion: Is Lebesgue integral w.r.t. counting measure the same thing as sum (on an arbitrary set)?


You could define $L$ to be the infimum of the set of all sums of finitely many terms of the series (not necessarily consecutive terms); and define $U$ to be the supremum of the same set. Then a series would be convergent if both $L$ and $U$ were finite and the sum would be $U+L$.

This would amount to defining a series to be convergent if and only if its positive-term subseries and negative-term subseries each converged (in the traditional sense) to a finite value. ($L$ would be the old-fashioned sum of the negative terms; $U$ of the positive.)

  • $\begingroup$ Interesting. Can you show this is different from absolute convergence (if it is)? $\endgroup$ – Ethan Bolker Oct 28 '18 at 17:50
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    $\begingroup$ I'm pretty sure it isn't different than absolute convergence. $\endgroup$ – paw88789 Oct 28 '18 at 17:52
  • $\begingroup$ This doesn't work at all, even for the nicely convergent sequence $a_n=2^{-n}$. You have $L=0$ and $U=2$ so it doesn't converge. For any conditionally convergent series you will have $L=-\infty, U=\infty$ $\endgroup$ – Ross Millikan Oct 28 '18 at 18:22
  • $\begingroup$ @RossMillikan So for $2^{-n}$ (for $n=0,1,2,...$ as you say, $U=2$ and $L=0$ so the sum would be $2$. I did have a mistake though, the sum should be $U+L$ not $U-L$, which I have noted in an edit. $\endgroup$ – paw88789 Oct 28 '18 at 18:55
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    $\begingroup$ I do wish you'd edit your original statement from "the sum would be $U - L$" to be "the sum would be $U + L$", instead of leaving the original incorrect statement in place and then adding a note which corrects it. $\endgroup$ – Tanner Swett Oct 28 '18 at 22:50

Just to give another point of view, you can see things in terms of integrals. The absolute summability Jack introduced is indeed the best way to go. In that case the sum is the integral of the function $n\mapsto a_n$ with respect to the counting measure on $\mathbb N$. The integral (just like a Lebesgue integral with respect to the Lebesgue measure) is only defined when the absolute value is integrable, which in this case means precisely that the usual sum $\sum_n|a_n|$ is finite.

That this integral with respect to counting measure is permutation invariant is fairly evident from the definition itself. The order structure on the set $\mathbb N$ is not used at all.


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