Summary: given an order and grouping of the summands, the sum of the set of integers, if convergent, converges to $0$.

Let...$$N=\sum_{n\in\Bbb{Z}}n=\sum_{n=-\infty}^\infty n$$ Rather than dismiss this as undefined, I would like to ask a simple question: what could the value of $N$ be?

Addition (of integers) is only commutative and associative for finitely many summands and this allows infinite summations to be reordered to obtain different values. It is a simple process to show this using the above summation...


$$N=(0+1)+(-1+2)+(-2+3)+(-3+4)+\cdots=1+1+1+1+\cdots=\text{n/a or }\infty$$

$$N=(0-1)+(1-2)+(2-3)+(3-4)+\cdots=-1-1-1-1-\cdots=\text{n/a or }-\infty$$

At first, I thought that it would be possible to rearrange the summands such that $N=k$ for any integer $k$, but then I noticed something. The sum only converges if, following simplification, it is the sum of finitely many nonzero terms; whenever this is the case, it seems that the sum must converge to zero. For instance...


Furthermore, this seems to be the case even if the sequence of summands is not order-isomorphic to the natural numbers...


(the order is as written, order type here is $\omega+3$)

What this suggests to me is that the only possible value (not counting $\pm\infty$) is $0$, and that this fact is intrinsic to the set of integers (with addition). If this is the case then my next question would be whether or not every countable summation has a countable (possibly empty) set of potential values, and how I would go about determining these values.


I had excluded this from the original question, because 1) I am still working on it and 2) I thought it would be more confusing than helpful. Since some clarification seems in order I am adding it anyway.

A couple of days ago, I asked myself a simple question: "what if, instead of thinking about infinite sums as limits, you think about them literally, as infinite successions of additions? How would you define such a thing?" After working on it for a bit I devised a precise (but complicated) way to define such infinite sums.

Let $<$ be a well ordering relation on a set $M$ and $f:M\to X$ for some set $X$ (with addition defined on $X$). The value of the infinite sum...

$$K=\sum_{m\in M} f(m)$$

...is defined precisely as follows:

  1. $\Sigma_<(E,f)=f({\min}_<E)+\Sigma_<(E\setminus\{{\min}_<E\},f)$ for any $E\subseteq M$.

(note, the existence of a minimum is guaranteed by $<$ being a well-order)

  1. $x+\Sigma_<(\emptyset,f)=x$ for any $x\in X$

In the case of the original question, $M=\Bbb{Z}$, $X=\Bbb{Z}$ and $f=\text{id}_\Bbb{Z}$. Thus...


...for some well-order $<$ on $\Bbb{Z}$.

Different groupings of summands can be defined accordingly:

For any set $X$, let $Pt(X)$ be the set of partitionings (or equivalence relations) on $X$

For any set $Y$, let $WO(Y)$ be the set of well-orders on $Y$.

For any set $Z$, let $<_Z$ be a well-order on $Z$.

For $P\in Pt(X)$, the grouping of summands can be expressed as $\Sigma_{<_P}(P,\Sigma_{C(A)}(A,f))$, where $C$ is a function assigning a well-ordering relation to each element of $P$.

So, for instance, if...






(Note that the choice of $C$ is irrelevant since each partition is finite)

Taking all this into consideration, my original question is whether or not the set of all summations of all partitions over the set of well orders on $Pt(\Bbb{Z})$ is $\{0\}$, or prove that...

$$\{\Sigma_{<_P}(P,\Sigma_{C(A)}(A,\text{id}_\Bbb{Z})):P\in Pt(\Bbb{Z})\land <_P\in WO(P)\land C:P\to\bigcup_{A\in P}WO(A)\}=\{0\}$$

I hope that this provides some clarification. Please keep in mind that I am still actively working on this and that the "theory" of well ordered sums, as it were, is less than a week old. While I have tried my best to make the notion I am describing precise, there are a lot of details that need to be covered. While I could probably be even more explicit, it could easily take a paper's worth of explanation (and I haven't even finished writing the paper yet).

Note: Technically, $\Sigma$ must be considered as a partial function whose domain includes only those well-ordering relations such that the sum converges. However, defining $\Sigma$ this way requires more work. Please keep in mind that I started this problem two days ago.

Note: There is a surjection from the set of well-orders on $\Bbb{Z}$ to the set of countable ordinals $\omega_1$. The summation can also be defined in terms of order-types and functions from a countable ordinal to $\Bbb{Z}$.

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    $\begingroup$ A plausible value would in fact be $0$ because the positive and negative numbers cancel out. I would choose this value, if I should choose one. $\endgroup$ – Peter Dec 11 '19 at 18:35
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    $\begingroup$ There are natural values to pick to define the sequence, and I too would choose $0$, as but the fact of the matter is that unless you have specifically defined what the sum means, it is simply undefined, since normal summation is not convergent in this case. $\endgroup$ – Jam Dec 11 '19 at 18:49
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    $\begingroup$ When a summation goes to different values depending on the order of addition or the grouping of terms, it is generally considered undefined unless the rules governing summation are prescribed. Consider $\sum (-1)^i$ which can go to $0,1,-1$ depending on whether and how the terms are grouped. Without the information on how the grouping is to be done, the sum is undefined. $\endgroup$ – Keith Backman Dec 11 '19 at 20:08
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    $\begingroup$ Even if you sum in the order $0 + 1 - 1 + 2 - 2 + 3 - 3 \cdots$, this isn't a convergent series in the usual sense, because the partial sums don't converge to zero: they oscillate without bound: $0, 1, 0, 2, 0, 3, 0, \ldots$. However, interpreting $\sum_{n = -\infty}^{\infty}$ as $\lim_{N \to \infty}\sum_{n=-N}^{N}$ is useful or even standard in many contexts, for example Fourier series. It's called the principal value of the series. $\endgroup$ – Bungo Dec 12 '19 at 3:14
  • $\begingroup$ If you stop thinking about sums as limits you're simply working out with a formal series or a close enough notion. $\endgroup$ – Asaf Karagila Dec 13 '19 at 18:40

I think if you're allowed to reorder and regroup at will, you can make the sum any integer you want. E.g., let's try to make it $3$. We go


At every stage, you take, among the numbers you haven't used yet, the one smallest in absolute value, and do what you have to do to cancel it. I think it's clear that you can never get stuck.

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  • $\begingroup$ Thank you! This was my original hypothesis, but I couldn't find an example (somehow, I didn't think to use groups of three or more summands). I'm still trying to figure out a rigorous proof. Intuitively, it does seems obvious just given one example (since it's easy to go from there) but then "maybe" this doesn't work for some specific integer. The main problem I'm encountering is keeping track of the numbers already added. I'ts easy enough to show that for any integer $n$, there exists some subset of integers whose sum is $n$, but I need to make sure that each integer occurs only once. $\endgroup$ – R. Burton Dec 15 '19 at 1:53
  • $\begingroup$ Let $n$ be the smallest number (in absolute value) not yet used. Say $n>0$ (the case $n<0$ uses the same idea). Let $m$ be the smallest number such that $m>n$ and $m$ hasn't been used and $-(m+n)$ hasn't been used. Then group $n$ with $m$ and $-(m+n)$. So you never get stuck. $\endgroup$ – Gerry Myerson Dec 15 '19 at 2:00

Let me answer with a question: what happens if you add up the negative and positive numbers at different speeds?

Wouldn't you also say that

$$ \sum_{n\in \mathbb{N}} n = \lim_{N \to \infty} \sum_{-N}^{N^2} n = \lim_{N \to \infty}\left(\frac{N^2(N^2+1)}{2} - \frac{N(N+1)}{2}\right) = +\infty ? $$

As it was mentioned in a comment, unless you assign a specific meaning to the sum, divergent is the best possible answer.

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$$N = 0 + (1-1) + (2-2)+(3-3)... = \sum_{n=-\infty}^{\infty}(n-n) = \sum_{n=-\infty}^{\infty}(0)=0$$ unless I’m missing something.

(NOTE: this answer was in response to the original question which has since been heavily edited to be much more general.)

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  • $\begingroup$ You are missing the term $0$, and your $\sum$ is from $n=1$ to $\infty$. But (added in your order) the answer is, as you say, $0$. $\endgroup$ – GEdgar Dec 11 '19 at 19:00
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    $\begingroup$ @TyJensen yes you are missing something... reordering the terms in a series is not a trivial question. You must understand that looking at this object as it was a normal sum leads many times to wrong conclusions. $\endgroup$ – PierreCarre Dec 11 '19 at 19:24

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