I'm reading about zeta-function regularization in physics and I have some mathematical doubt. I understand that, since a sum of infinite terms is not well defined in a field, a series that is considered divergent in the usual meaning can have a ''value'', defined in some less conventional way.

The zeta-function regularization is one of such ways that, as an example, assign the value $-\frac{1}{12}$ to the infinite series $S=1+2+3+4+...$ using the fact that the zeta-function $\zeta(s)$ is the analytic continuation of the series $\sum_{n=1}^\infty n^{-s}$, and $\zeta(-1)=-\frac{1}{12}$.

But how we can be sure that there does not exists other possible ''regularizations'' that gives different values to the same series?

And, if more than one value does exist, there is some criterion to select between them? Or the zeta-function regularization is preferred only for physical motivations (because the experiments confirm its values)?

On the web I've found a lot of posts less or more reliable about this topic, someone knows a reference to a well defined axiomatic approach to the problem of the value of divergent series?

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    $\begingroup$ Well, when regularizing an infinite sum, we usually try to enforce a few "rules", as seen here. The less rules you follow, the likelier you are to produce inconsistent results, though if you follow the rules... you will almost always produce one specific value. :-) Works for rearrangement of conditionally convergent series too. $\endgroup$ – Simply Beautiful Art Mar 3 '17 at 22:07
  • $\begingroup$ In the typical sense of a summation, no. Realize that I can separate each term into $1 + 1 + 1 + 1 + ...$ and by performing algebra, I can get this to sum to anything I could ever want. The summation is divergent in the usual sense, and isn't useful to any definition summations are typically useful for. Though I admit I don't know much about the rules that beautiful art posted above, most people who push $1 + 2 + 3 + 4...$ aren't pushing it from a knowledgeable point of view. $\endgroup$ – Kaynex Mar 3 '17 at 22:11
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    $\begingroup$ Different regularizations may not agree. For example, the assignment $\sum_{n=1}^{\infty} n = \infty$ is perfectly acceptable for any summability method that exhibits positivity, such as ordinary summbability, Cesaro summability or Abel summability. On the other hand, zeta regularization is a kind of consistent way of extracting meaningful value, at least in the sense described in Terry's blog. $\endgroup$ – Sangchul Lee Mar 4 '17 at 0:50
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    $\begingroup$ @SimplyBeautifulArt, I may indeed be ignorant at what physicists call a regularization. I googled and found this interesting post from physics SE. In this question is presented the following unusual regularization $$ \lim_{s \to 0^+} \left[ \text{analytic continuation of } \sum_{n=1}^{\infty} \frac{n}{(n+\alpha)^s} \right] = \frac{\alpha^2}{2} - \frac{1}{12}. $$ This would be a better example for my claim. $\endgroup$ – Sangchul Lee Mar 4 '17 at 1:00
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    $\begingroup$ Many regularizations seem to introduce a new parameter or "regulator" to make a sum finite, which is then removed at the end in some subtraction or limiting process. A lot of regularizations seem to yield the same result for the series in question (the "magic" $-\frac{1}{12}$); whether this is something deep or it's just self-fulfilling (this seems to be the "benchmark" series in testing whether a regularization "works") I am not sure. As soon as you allow two parameter regularizations though (why not?) I think all bets are off and different "regularizations" can yield different "results". $\endgroup$ – AloneAndConfused Mar 4 '17 at 11:36

Our problem is: Given a sequence $(a_n)_{n=0}^\infty$ of numbers, assign a "sum" to it, i..e, give a meaning to the pixel pattern $\sum_{n=0}^\infty a_n$. This is easy the sequence of partial sums converges as we can simply assign the limit of these. This does not define $\sum_n a_n$ for all given sequences $(a_n)_n$, but only to a certain subset $S$ of the set of all sequences. But we have a collection of nice rules:

  • $S$ is a vector space and $\sum$ is linear, that is
    • If $(a_n)_n,(b_n)_n\in S$, then $(a_n+b_n)_n\in S$ and $\sum(a_n+b_n)=\sum a_n+\sum b_n$
    • If $(a_n)\in S$ and $c\in\Bbb C$, then $(ca_n)_n\in S$ and $\sum ca_n=c\sum a_n$.
  • $S$ is closed under adding/dropping/modifiying finitely many terms, which boils down to
    • $(a_n)_n\in S$ if and only if $(a_{n+1})_n\in S$. In this case $\sum a_{n}=a_0+\sum a_{n+1}$

Regularization is the attempt to enlarge $S$ in a useful way. There are many ways to do so, and in particular if one ignores the above nice rules, one can do it almost arbitrarily (using the Axiom of Choice, perhaps). On the other hand, if one wants to have permanence of the above rules (or perhaps others), then quite often the extension is uniquely determined.

For example, if we want to enlarge $S$ in such a way that it contains $(2^n)_n=(1,2,4,8,\ldots)$ and we assign (by any computational method at all) a value $\sum 2^n=c$, then $S$ must also contain $(2\cdot 2^n)_n=(2,4,8,16,\ldots)$ and $\sum 2^{n+1}=2c$, and $S$ must contiain this prepended with $1$ - which is again the original sequence - and so we obtain the equality $2c+1=c$. Therefore, if we want to assign a value to $\sum 2^n$ and want permanence of the above rule, we must agree to set $\sum 2^n=-1$. Similarly, it follows that $(1)_n$ cannot be $\in S$ because $c+1=c$ has no solution.

But can we extend $S$ to contain $(n)_n$? If so, then also $(n+1)_n\in S$ and their difference $(1)_n\in S$ - which we have just seen is impossible. Thus any attempt to consistently assign a value to $1+2+3+4+\ldots$ must drop one of the very reasonable rules listed above. But with these rules as goalposts removed, one would first have to agree what constitutes a "valid" extension of summation before one can make statements about whether a certain value (if it is assigned at all) is necessarily correct (as we did above for $\sum 2^n$).

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  • $\begingroup$ Thank you @Hagen. So the zeta-function regularization is not better than other possibilities from a mathematical point of view. It is ''preferred'' only because it seems that it works better experimentally, in the sense that there are experiments whose results are compatible with such kind of regularization. It is a case in wich physics select one of the possible mathematical ways. $\endgroup$ – Emilio Novati Jun 24 '17 at 12:31
  • $\begingroup$ @EmilioNovati "So the zeta-function regularization is not better than other possibilities" How did you get that conclusion from this answer? $\endgroup$ – user76284 Jun 30 '19 at 2:45
  • $\begingroup$ Awful answer... $\endgroup$ – Anixx Aug 29 '19 at 19:57

I'll go a bit further than Hagen Von Eitzen's (very good) answer.

I had considered the general case as he does, to see whether $1+2+...= -1/12$ had any "deep" content.

So I'll give here the results I found (which do not, if I remember correctly, require the axiom of choice), and if you are interested I can provide proofs.

Call admissible space a sub-vector space of $K^{\Bbb{N}}$ for $K=\Bbb{R}, \Bbb{C}$ that contains all summable sequences and is closed under $(a_n)_n \to (0,a_0,a_1,....)$.

If $H$ is an admissible space, and if $T$ is a linear form on $H$, say that $T$ is a supersummation if and only if :

  1. it extends the sum for summable sequences, and
  2. $T((0,a_0,a_1,\ldots)) = T((a_n)_n)$, ie if appending $0$ to the front of a sequence does not change the ‘sum’.

Say that a supersummation $T$ on the admissible space $H$ is proper if and only if $T$ is the unique supersummation on $H$.

Now one has the following results :

  • Assume $H$ is an admissible space, with a proper supersummation $T$ and $u$ is a given fixed sequence. Then $H$ is contained in an admissible space containing $u$ with a proper supersummation on it, if and only if $u$ satisfies a linear induction relation, up to an $H$-term. The second condition reads, more precisely: there exist $k\in\Bbb{N}$, $a_0,\ldots,a_{k-1}$ and $h\in H$ such that $u_{n+k} = \displaystyle\sum_{i=0}^{k-1} a_i u_{n+i} + h_n$ for all $n\in\Bbb{N}$.

  • There exists a (necessarily unique) admissible space $H$ satisfying the following properties: 1. $H$ has a proper supersummation; 2. no admissible space properly containing $H$ has a proper supersummation; moreover 3. every admissible space with a proper supersummation, $(H’,T’)$, is contained in $(H,T)$, ie, $H’\subseteq H$ und $T’=T\mid H'$.

This second result is very interesting as it can be read ‘If you want to define a generalization of infinite sums, there is only one way to do it right, and the way of doing it is, in a sense, absolute’. And, as Hagen Von Eitzen shows, $H$ cannot contain $(n)_n$ but it does contain $(2^n)_n$, and the value of the supersummation of $(2^n)$ is $-1$.

EDIT: I'm adding to this answer a short "paper" that I had written about this topic. A few remarks : 1. This is in french (because I'm french), so until I trnalsate it (if I translate it), only those who read french can understand it. 2. I wrote it at the beginning of my second year post-highschool, so there are some things that I could have done differently and really more easily -don't be surprised if there are calculations that could be avoided, or trivial arguments that are developed etc. 3. At some point I use Zorn's lemma, but as those who read it can see, it's not necessary. I simply use it to get the existence of a maximal element, but I prove its existence in another way later: I'm only using it to show what the maximal element might look like, so that I can better pursue the search for said element. With this in mind, here's the "paper"

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  • $\begingroup$ The two final results are indubitably interesting (If you have some reference where I can see the proof I'll be grateful). All this confirms that the zeta-function method is preferred only for physical reasons, and cannot be preferred for a mathematical reason, because it is incompatible with such '' maximally supersummable space'', that does not contain the sequence $(n)_n$. do I well understand? $\endgroup$ – Emilio Novati Jun 24 '17 at 13:19
  • $\begingroup$ Yes you understand correctly. Unfortunately, I cannot provide any reference because I proved these results myself (I also had them checked by my teacher so the chance that they're actually wrong is reasonably low) and had not found any previous reference. If you speak french I can reference a video where the narrator gives some of the main ideas of this result (which I only generalized and led to their end). If you don't and wish to see proofs, tell me and I'll translate my work for you and post it somewhere accessible $\endgroup$ – Maxime Ramzi Jun 24 '17 at 13:34
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    $\begingroup$ merci! je comprend un petit peu le francais :) $\endgroup$ – Emilio Novati Jun 24 '17 at 13:43
  • $\begingroup$ I've added a link in my answer to the "paper" I wrote about this. As of now, it is in french. I can translate it if I have time and if I'm asked. $\endgroup$ – Maxime Ramzi Jun 24 '17 at 14:18
  • $\begingroup$ merci beaucoup. $\endgroup$ – Emilio Novati Jun 24 '17 at 14:32

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