Assume I'm unhappy with the Riemann Rearrangement Theorem. How else can I define a series?

Question

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.

Answer 1

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 $\left|\sum' a_{2n}\right|=+\infty.$

Answer 2

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.)

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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)?

Answer 3

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.)

Answer 4

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.