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I recently asked this question about building an intuition for summations of divergent series. The question was based on the fact I had recently learned that $1 + 2 + 3 + ... = \frac{-1}{12}$. My question was about how to interpret this result given that the summation $1 + 2 + 3 + ...$ diverges.

This question was closed as a duplicate of this earlier question, which seemed to ask how it would be possible at all for $1 + 2 + 3 + ... \frac{-1}{2}$ given that the sum diverges. I think that this question is related, but isn't a duplicate. The question I asked was primarily about how to think about the "meaning" of assigning a sum to a divergent series. Specifically, the core of my question was this part:

Right now, I have two different hypotheses about how to understand these results together simultaneously:

  1. There are two different definitions of what an infinite summation "means." One uses limits of finite summations, and the other uses complex analysis. Since these definitions aren't identical to one another, it's not surprising that they predict different results in this case, and the disparity arises because there are two different definitions that happen to use similar notation to describe their results.

  2. Just because the partial sums of the first 0, 1, 2, 3, 4, 5, ... etc. terms of the series don't converge to a value doesn't mean that the sum of all infinitely many terms doesn't have a value. The series diverges because the finite partial sums don't converge to anything, but the infinite summation really is indeed $\frac{-1}{12}$.

Are either of these hypotheses correct? Or am I off-base here? I'm hoping to learn how to think about results like these, and if there's some bigger picture that everything fits into I'd appreciate more information about it.

Given that I've just learned about summing divergent series, it's quite possible that the linked question really is a duplicate and I just don't have enough of a math background to understand why. If that's the case, then I apologize for asking a duplicate question. If not, though, would it be possible to reopen my question?

Thanks!

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  • $\begingroup$ There were way to many questions on this topics recently; I'm sure you can find answer to your question there. On the top of it, you asked two question on this topic in 2 minutes, right? This is not exactly encouraged... $\endgroup$
    – Grigory M
    Commented Jan 12, 2014 at 18:13
  • $\begingroup$ @GrigoryM I asked those questions separately because I was under the impression they covered different topics. The first was on a specific flaw in the reasoning of a proof, while the second was supposed to be about intuiting sums of divergent series. I apologize if that wasn't clear. $\endgroup$
    – templatetypedef
    Commented Jan 12, 2014 at 18:23
  • $\begingroup$ Well, in general it's nothing wrong in asking two related but different questions (although it usually makes sense to ask first one, think about received answers and only then ask second one) — but in the situation when questions on $1+2+3=-1/12$ are asked constantly (more than 10 questions in last 2 days, I think; many of them literally duplicates)... $\endgroup$
    – Grigory M
    Commented Jan 12, 2014 at 18:27
  • $\begingroup$ @GrigoryM To your first comment, I agree that asking two questions on similar topics (especially when the first one might have produced answers that cleared up the second) is not a good idea, and templatetypedef should think twice before doing that. But I think that has no bearing on the current matter of "is this question templatetypedef asked a duplicate of this other question they didn't?". Re: second comment, I think not all questions spurred by that equation need be the same, though, and "there have been a lot of related questions recently" seems independent from the issue at hand. $\endgroup$
    – Mark S.
    Commented Jan 12, 2014 at 18:30
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    $\begingroup$ ...Finally, shifting from meta-discussion to mathematics I really hope you will find two questions linked below ('infinity...' and 'analytic continuation...') of some use. $\endgroup$
    – Grigory M
    Commented Jan 12, 2014 at 18:33

2 Answers 2

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Both the difference between limit of partial sums and analytic continuation (your 1.) and whether $1+2+\ldots$ 'really is' $-\frac1{12}$ (your 2.) have been extensively discussed in Why does $1+2+3+\cdots = -\frac{1}{12}$? (of which your question is marked as duplicate) and in many questions linked there (see in particular — much more accessible, perhaps — $\infty = -1 $ paradox and Analytic continuation- Easy explanation?).

I don't see how this question is not a duplicate.

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  • $\begingroup$ I don't think that question, whose primary answers are in-depth calculations/justifications of the value $-1/12$ answers the less technical question "would I be right in saying that there are different definitions of sum?". In fact, the highest voted answers to that other question may obscure this issue, by making it look like there's essentially one way to sum it, which yields $-1/12$. Your summary of 2. differs significantly from how I read it in the original question, although I may very well have misread it/your interpretation may agree with the consensus. $\endgroup$
    – Mark S.
    Commented Jan 12, 2014 at 18:35
  • $\begingroup$ (Re: Your summary of 2. differs significantly) well, the infinite summation really is indeed −1/12 is a verbatim quote from the (meta) post we're discussing $\endgroup$
    – Grigory M
    Commented Jan 12, 2014 at 18:37
  • $\begingroup$ You're absolutely right on that point. I suppose I was confusing my feeling that the top answers to that linked question didn't really answer either question with the matter of what either question was. $\endgroup$
    – Mark S.
    Commented Jan 12, 2014 at 18:40
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    $\begingroup$ (As a side note) it would be nice, I think, if someone could post some less technical answer to math.stackexchange.com/q/39802 — explaining problems with divergence, existence of various methods of summation and basic idea of an. cont... $\endgroup$
    – Grigory M
    Commented Jan 12, 2014 at 18:43
  • $\begingroup$ (I mean, Matt E.'s answer there is (as always) very good, IMO; and these less technical things are explained elsewhere — but since there a lot of people coming to this question now...) $\endgroup$
    – Grigory M
    Commented Jan 12, 2014 at 18:45
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I agree that this isn't really a duplicate. Edit: It seems like the original question was are about the nature of definitions of infinite summation, while the top answers at the other one are about calculations that show that under essentially one definition (or more), it comes out to $-1/12$, while not addressing the seeming contradiction with the divergence of the series.

The closest thing to an obvious answer (as opposed to one obscured by details you were not asking about) to your question on that other page may be the low-vote answer linking to the wikipedia page for Ramanajun summation, which internally links to the page for divergent series, which perhaps indirectly answers your question.

(Incidentally, as I don't have 3000 reputation, I can't cast a reopen vote. You may want to post an answer at Requests for Reopen & Undeletion Votes, etc. (volume 10/2012 - 12/2014) linking to this meta question.)

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  • $\begingroup$ Fundamentally, this may come down to an issue of what it means to be duplicates. If questions are duplicates when one could support an answer covering both, then maybe they are. I think the other question could have garnered an answer which would answer templatetypedef's questions (because I agree there is some overlap in topic), but I don't believe it did, and since it has some extremely-high-voted answers, so I don't think it's likely to serve as a good resource in the future for people asking templatetypedef's question. $\endgroup$
    – Mark S.
    Commented Jan 12, 2014 at 18:48
  • $\begingroup$ I concur, and would like to add that just like proof-verification isn't considered a duplicate of similar questions, there's a case to be made that seeking to reconcile what seems like conflicting interpretations shouldn't be considered as such, either. The process is highly individual. $\endgroup$
    – Jonathan Y.
    Commented Jan 20, 2014 at 8:00

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