I learned about the Riemann rearrangement theorem recently and I'm trying to develop an intuition as to why commutativity breaks down for conditionally convergent series.

I understand the technique used in the theorem, but it just seems really odd that commutativity breaks down to me. It doesn't happen for other properties -- associativity holds for convergent series and commutativity holds for absolutely convergent series. What makes this particular property for this particular kind of convergent series "special" in this way?

I'm aware of this question: why does commutativity of addition fail for infinite sums? but the answers haven't been helpful for me. JiK's answer is "why would it?", and goes on to talk about why you can't apply rules to infinite series, but this seems erroneous because associativity holds for convergent series and commutativity holds for absolutely convergent series. Fly by Night just explains conditionally vs absolutely convergent, josh314 says it applies to finite sums only which isn't true, and Denis just explains the theorem again, and Barry Cipra seems to have a similar kind of argument as JiK's and problematic for similar reasons.

Is there a good way to understand why this is happening intuitively? or is this the wrong way to think about it and I should just accept that it's happening even though it's unintuitive? It's hard for me to just let it go without an intution because it seems like math starts to "break" here.. the theorem is sound but this property no longer holds in this case, which is really strange to me

Does anyone know of any resources that go into depth on this kind of question?

  • $\begingroup$ There is exactly one way to acquire intuition: lots of exercise. You don't get it by asking philosophical questions in mathematical places. $\endgroup$
    – user436658
    Commented Feb 20, 2018 at 22:40
  • $\begingroup$ Does this help? math.stackexchange.com/questions/2655704/… $\endgroup$
    – SK19
    Commented Feb 20, 2018 at 22:43
  • $\begingroup$ Oh, and make sure to read Euler about this topic. For example, he concluded that $\sum(-1)^k=\frac{1}{2}$. This is not oldschool convergence, but I have seen the reasoning why assigning this value to this divergent series can make sense. $\endgroup$
    – SK19
    Commented Feb 20, 2018 at 22:44
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    $\begingroup$ I don't think you mean "commutativity" when you say "commutativity". The meaning of an operation $\oplus$ being "commutative" is that $a \oplus b = b \oplus a$ for all $a,b$. So not only does the term "commutativity" not apply to infinite sums, it only applies directly to binary sums. When you say that the commutative property of addition does not hold for conditionally convergent sums, ask yourself what precisely you mean. $\endgroup$ Commented Feb 20, 2018 at 22:46
  • 2
    $\begingroup$ @Carl Mummert Together with assoacivity commutativity can be transferred onto arbitrary finite sum. And certainly the sum of conditionally convergent series will not change if only finite number of terms is permuted. Here is the point where the "intuition" fails. $\endgroup$
    – user
    Commented Feb 20, 2018 at 22:56

4 Answers 4


The issue here is not really "commutativity". The question is, why can rearranging the terms of a series cause it to converge to a different value?

The answer is that the definition of the limit of a series uses the order of the terms to define the sequence of partial sums, and then takes the limit of that sequence. If you rearrange the terms, you get a different sequence of partial sums, and there is no reason the new sequence of partial sums needs to converge to the same limit as the old one.

There is a kind of convergence that does not use the order of the set of terms being added: unordered summation. If $A$ is a set of real numbers, then $\sum_{i \in A} i$ is defined as the limit of $\sum_{i \in F} i$ over the net of finite subsets $F$ of $A$. Unordered summation has no "conditional convergence": we have that $\sum_{i \in A} i$ converges if and only if $\sum_{i \in A} |i|$ converges.

This helps show how the phenomenon of conditional convergence is related to the fact that rearrangements of a series can converge to different values: it is because of the particular way that the limit is taken in the definition of series convergence.

The same phenomenon happens in integration. An improper integral such as $\int_1^\infty f(x)\,dx$ converges if and only if $\lim_{k \to \infty} \int_1^k f(x)\,dx$ converges. So "rearranging the values" of the function $f(x)$ in particularly simple ways can cause the improper integral to converge to a different value, if the integral is conditionally convergent.

For example, for each series $\sum a_n$ there is a piecewise constant function $f(x)$ so that $\int_1^\infty f(x)\,dx$ is exactly $\sum a_n$. Namely, $f(x)$ has value $a_i$ on $[i, i+1)$. Rearranging the pieces of the graph of $f(x)$ is the same as rearranging the series $a_n$. So the improper Riemann integral does not just compute the area between the $x$ axis and the graph of $f(x)$, and the series $\sum a_n$ does not just add up the terms. The definitions of convergence rely fundamentally on an underlying order relation to take a limit.

The alternative kind of integration known as Lebesgue integration is more like unordered summation. There is no conditional convergence in Lebesgue integration, and rearranging function values (via a measure preserving transformation) cannot cause a convergent Lebesgue integral to converge to a different value.

  • $\begingroup$ I'm not sure the part about integrals makes much sense. It would be easy to come up with examples of functions $f$ and $g$ where $g$ is bijective, but $\int f\circ g \ne \int f$ in both the Riemann and Lebesgue integrals. Easy example: $f(x) = \frac{1}{1+x^2}$, $g(x) = 2x$, domain is $\mathbb{R}$. $\endgroup$ Commented Feb 20, 2018 at 23:12
  • $\begingroup$ @Daniel Schepler: I was not trying to make a formal definition of "rearrange the values", but I hope the concrete example of a function corresponding to a series shows the kind of example I was talking about. As a formally correct theorem, if we consider the set of intervals $S = \{[i, i+1) : i \in \mathbb{N}\}$, a Lebesgue integral $\int_1^\infty f(x)\,dx$ that converges gives the same value if we change $f$ by permuting $S$. $\endgroup$ Commented Feb 20, 2018 at 23:14
  • $\begingroup$ Here is a set of lecture notes that shows the right notion of "rearranging the values of $f$" for Lebesgue integration - the right notion is to use a measure preserving transformation. statslab.cam.ac.uk/~james/Lectures/pm20.pdf In the case where our domain is the natural numbers, every rearrangement is measure preserving (with the counting measure) $\endgroup$ Commented Feb 20, 2018 at 23:21
  • $\begingroup$ Thanks, this makes a lot of sense to me. So the way I understand it is: 1. Conditional convergent series says that order matters, so by extension of this, commutativity of addition can't hold. Absolute convergent series says order doesn't matter which is why commutativity still holds there, and 2. The definition of a given convergent series defines it's order, so if you rearrange anything in it then you're defining a different series rather than manipulating the same one $\endgroup$
    – risto
    Commented Feb 21, 2018 at 12:28
  • $\begingroup$ @CarlMummert is this the right intuition to have? My thinking is that even though conditionally convergent series doesn't explicitly say that order matters, it's implied because absolute convergence doesn't hold $\endgroup$
    – risto
    Commented Feb 21, 2018 at 13:12

"commutativity" is normally defined for two elements (addition is at root a binary operation) and extended to any finite number of elements.

The limit operation for infinite sums is defined without reference to the commutative property. There is no reason therefore for the limit operation to respect commutativity.

In the case of conditional convergence I think it is intuitive to see that (a) the sum of the positive terms alone is unbounded; (b) the sum of the negative terms alone is unbounded; (c) if I can change the order however I like I can push the negative terms as far down the order as I care to determine, and because the sum of the positives is unbounded, I can overwhelm each negative by positives; (d) I can likewise push the positives as far as I care down the order, and overwhelm the positives; hence (e) different orders have different behaviours; and (f) convergence and limits depend on order.

To add a second thought, which may be helpful. I can change the order of a finite number of summands in an infinite sum without changing convergence or limit properties. Why? Because then there is an $N$ after which nothing has changed, and the sum up to $N$ is also the same as it originally was.

So we could say that the (finite) commutative property does still apply.


I think the 'right' intuition you should have here is:

Commutativity of addition only guarantees that a sum is preserved after finitely many terms are swapped.

It so happens that the above still 'holds' for an infinite sum, since it is defined as a limit of partial sums and hence from a certain point onwards the swapped terms are all included in the partial sum.

It does not imply that you can use an arbitrary permutation of the terms in the infinite series, as there are obviously permutations that cannot be expressed as a finite composition of swaps.


Consider a series that is convergent but absolutely divergent, such as $$\left\lbrace u_n= \frac{(-1)^n}{n}\right \rbrace_{n\geq 1}$$

Consider the positive and negative parts: $$u_n^+=u_n\qquad \text{if}\ \ u_n>0\qquad\qquad u_n^+=0\qquad\ \text{otherwise}$$ and $$u_n^-=u_n\qquad \text{if}\ \ u_n<0\qquad\qquad u_n^-=0\qquad\ \text{otherwise}$$ Claim: $\sum u_n^+$ diverges, and so does $\sum u_n^-$.

Proof: At least one of them has to diverge, since $\sum |u_n|$ diverges. But then the other one has to diverge as well, otherwise $\sum u_n$ would diverge.

So our series contains enough positive numbers to tend to infinity and enough negative numbers to tend to infinity in the other direction!

Claim 2: For any $\varepsilon >0$, $u_n$ contains only finitely many terms of absolute value larger than $\varepsilon$.

Proof: This follows directly from the fact that $\sum u_n$ converges.

Intuitive proof of Riemann's theorem: By Claim 1, one can take a sum of some suitably chosen $u_n$ so as to get close to any target $L\in \Bbb R$, and by Claim 2 one can decide to pick the $u_n$ in a suitable order so as to stay close to $L$ (and even tend to $L$).

One can also decide to pick the elements in an order that will make the sum diverge. The trick is to realise that a bijection $\Bbb N\to \Bbb N$ can really mess badly with the intuitive order. You could take the two bigger positive $u_n$, then a negative, then the next $4$ bigger positive $u_n$, then the next negative, then the next $8$ positive $u_n$... that would be a bijection. It would not necessarily diverge though. Here is one that surely does: take enough positive $u_n$ to get over $10$, then one negative, then enough positive to get over $100$, then one negative, ... this is still a perfectly fine reordering!


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