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If we agree to define $S=\lim_{n\to \infty}\sum_0^na_n$,then, only convergent series can have meaningful $S$. But if we decide to define $S$, to be the Caesaro mean, now we can also assign a real number to $S$ for some divergent series as well. My question is such relaxation of the definition frown upon in the mathematical community? The reason I ask is I have read several blog posts where authors are adamant that only convergent series are summable and only summmability in the ordinary sense is useful. In those articles, they usually point out the Ramanujan sum as an example of getting nonsensical answers. They especially have huge problems with getting results by analytical continuation of Rieman zeta function. To me Caesaro summation and Abelian summation are natural extensions of the summation in ordinary sense. Where do you draw the line?

P.S. I am very new to this topic and mathematics in general so please bear with me for my naive question.

curiosity
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    Mathematicians don't have any inherent problem with Cesaro sums, regularizations, etc.; they're well-defined and useful. The problems are with treating these operations on infinite sequences as the same as the ordinary sum (i.e., the limit of partial sums) and blithely applying the same calculus to them. There's a meaningful and useful way to define a function on some infinite sequences that takes the value $-1/12$ at $(1, 1, 1, \dots)$, to take the inevitable example, but it is absolutely not true in any sense that the sum $1 + 1 + \cdots$ converges to $-1/12$. – anomaly Mar 14 '24 at 12:38
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    I have read several blog posts --- Are you sure the authors have much awareness of the vast subject of mathematics? There are thousands of papers involving summability methods and dozens of books (my comment here gives 3 books between 1913 and 1916 alone), and for some areas (e.g. trigonometric series -- see Zygmund's treatise) summability methods are essential tools. – Dave L. Renfro Mar 14 '24 at 12:42
  • @DaveL.Renfro Here is the link. Did I misinterpret what the author was saying? https://math.ucr.edu/home/baez/physics/General/summingNaturals.html – curiosity Mar 14 '24 at 13:10
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    @anomaly, I agree totally with the sentiment of your comment. But a technical quibble: isn't the inevitable example that $(1, 2, 3, \dotsc)$ "sums" to $-1/12$? And I think this is in the sense of eg Ramanujan summation but not Cesàro summation. I believe the Cesàro sum of $(1, 1, 1, \dotsc)$ is $1$ and its Ramanujan sum is $1/2$. – Izaak van Dongen Mar 14 '24 at 13:20
  • @IzaakvanDongen: Gosh, of course, thank you. It's past the edit window for the comment, but that should definitely be $1 + 2 + \cdots$. (For the sake of the OP: The idea is to regularize the sum to wind up with $\zeta(-1) = -1/12$, since the sum $\zeta(s) = \sum n^{-s}$ has a meromorphic continuation to $\mathbb{C}$.) – anomaly Mar 14 '24 at 14:05
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    @anomaly would you fix the typo in your comment and write it up as an answer. Thank you all so much. – curiosity Mar 14 '24 at 14:42
  • Here is the link --- I've skimmed over the blog entry you linked to and didn't see anything remotely close to "authors are adamant that only convergent series are summable and only summability in the ordinary sense is useful". – Dave L. Renfro Mar 14 '24 at 14:46
  • Do you want to know the opinion of the mathematical community or how we (the users here) think about this ? The value we get depends on the function we want to extend , and this is the problem and the reason I do not consider those assignments as useful. I doubt that the function and therefore the value is unique. – Peter Mar 14 '24 at 14:51
  • Anyway , it is important to distinguish between the (divergent) sum and the assignment. Something like $1+2+3+\cdots=-1/12$ is NOT right in "some sense" , the -1/12 is only an assignment due to an extension , not anymore a sum. – Peter Mar 14 '24 at 14:52
  • @DaveL.Renfro, Perhaps I misunderstood the article. I thought the author was rejecting Caesaro sum in the paragraph following equation (12). – curiosity Mar 14 '24 at 14:53
  • @Peter. OK, thanks. Is Caesaro summation better than this example or is it in the same category? – curiosity Mar 14 '24 at 14:56
  • The most problematic detail of those assignments is not that they are used in math (this is OK , as long as one tells the reader "what is going on") , but the distributed myth that such assignments have applications in theoretical physics. Sad that this myth has been cemented. I heard this is the foundation for the string theory , in which case nothing more has to be said. – Peter Mar 14 '24 at 14:58
  • @curiosity As far as I understood , Cesaro averages and can be applied in oscillating sums , like as $1-1+1-1+1-1+\cdots$ , which we can this way assign 1/2 – Peter Mar 14 '24 at 15:00
  • @curiosity: Sure, happy to. If you'd like more details, let me know; the original comment was just a comment. – anomaly Mar 14 '24 at 15:50
  • I think the intended meaning behind "the Cesaro sum does not correspond to anything about the real world" is that in successively counting/adding things, Cesaro summation isn't helpful. But I could similarly say that non-real complex numbers do not corresponding to anything about the real world if we're counting the number of trees in a forest. My impression is that the author is mostly discussing an unwarranted (or at least, a not appropriately supported) involvement of certain summability methods in certain physics arguments (but this is mostly a hunch). – Dave L. Renfro Mar 14 '24 at 15:55
  • @DaveL.Renfro Thanks, your comment answers what was bothering me. I thought I understood the idea behind the Caesaro sum and then I read the third paragraph following equation (12) and it pulled the rug under me. "The analogy is used that the square root of 2 doesn't exist in the rational numbers, so we extend those numbers to become the real numbers," – curiosity Mar 14 '24 at 16:43

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Mathematicians don't have any inherent problem with Cesaro sums, regularizations, etc.; they're well-defined and useful. Cesaro sums pop up in Fourier analysis, for example, and Tauberian theorems in number theory have at least the same spirit as the sort of regularization that comes up in dealing with these divergent sums.

The problems are with treating these operations on infinite sequences as the same as the ordinary sum (i.e., the limit of partial sums) and blithely applying the same calculus to them. There's a meaningful and useful way to define a function on some infinite sequences that takes the value $−1/12$ at $(a_n) = (1, 2, 3, \dots)$, to take the inevitable example, but it's absolutely not true in any sense that the sum $1 + 2 + 3 + \cdots$ converges to $-1/12$.

anomaly
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You can argue that Ceasaro and Abelian summation are the most natural extension of the standard infinite summation but that doesn't make a summable divergent series convergent. There are a lot of significant properties that a convergent series would have and a divergent series wouldn't so it's best to keep thing precise to their meaning and avoid ambiguity, which can later on lead to contradiction if you're not careful. You can just call them Caesaro or Abel summable.

An example is grouping consecutive terms in a series. For a convergent series, you can do that however you want and the result wouldn't change. But for a Caesaro summable series, that isn't true. $1+(-1+1)+... = 1+0+0+.. = 1$ Whereas $(1-1) + (1-1) + ... = 0+0+... = 0$

ioveri
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  • Yes, but is it easy to reject the summation method in your example because associativity does not need to hold for infinite series. Also, rearranging terms are not allowed (Riemann rearrangement theorem). But Cesaro and Abelian summation methods do not use any such illegal operations and are regular, linear and stable. So I agree with you in saying they are the most natural extension. Thank you for your thought ful answer. – curiosity Mar 18 '24 at 13:07
  • Grouping consecutive terms does not require associativity because you're not moving the terms around, you're just sum them in groups. The sequence of partial sums of resulting series is a subsequence of the original one. Hence if the original one is convergent then the new series also converges to the same value. In my example the subsequences are $a_1,a_3,a_5,...$ and $a_2,a_4,a_6,...$ respectively – ioveri Mar 18 '24 at 19:09
  • Hmmm....associativity is not moving the terms around. Associativity is changing the order of the operation, hence grouping consecutive terms like you are doing, right? Am I missing something? – curiosity Mar 19 '24 at 11:46
  • Sorry, I meant commutativity. Riemann rearrangement theorem concerns commutativity, not associativity. The "rearrangement" corresponds to permutation of terms, hence grouping terms isn't a violation. Associativity holds for convergent series. The proof is simple: the sequence of partial sum of the new series is a subsequence of the sequence of partial sums of the old one, hence it must be convergent and share the same limit.

    And while Caesaro summation does not use grouping, and is indeed linear, regular stable, associativity does not hold for Caesaro summable sum

     – ioveri Mar 19 '24 at 12:09