A while ago, there was a great hype about the “identity” $$\sum_{n=1}^{\infty} n = -\frac{1}{12}.$$

Apart from some series manipulations where the validity seems to be at least questionable, the derivation of this always goes through the zeta function:

Where the series converges, the zeta function is defined by $$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s}$$ and outside that range by analytic continuation. And it turns out that inserting $s=-1$ formally results in $$\zeta(-1) = -\frac{1}{12} = \sum_{n=1}^\infty n$$

However looking at the series in isolation, there is no indication that the zeta function should be chosen.

An obvious way to get an analytic function that at one point gives the sum of all natural numbers is $$f(x) = \sum_{n=1}^{\infty} nx^n$$ at $x=1$, however (not surprisingly) that function diverges at $1$.

Therefore my question:

Is it possible to get another finite value for the series by analytic continuation of another series?

Concretely, do there exist continuous functions $f_1, f_2, f_3, \ldots$ such that

  • On some non-empty open subset $S$ of $\mathbb C$, $f(x)=\sum_{n=1}^\infty f_n(x)$ converges to an analytic function.

  • At some point $x_0$, $f_n(x_0) = n$ for all positive integers $n$.

  • The analytic continuation of $f$ is well defined and finite at $x_0$.

  • $f(x_0) \ne -1/12$

What if we demand the functions $f_n$ to be analytic rather than just continuous?

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    $\begingroup$ is not "questionable" , the identity is just false. A possibility to write correctly the idea behind this "identity" is writing $\sum_{n=0}^{\mathfrak R}n=-\frac1{12}$ where the $\mathfrak R$ means that we used the Ramanujan summation method to give a value to the divergent series $\endgroup$
    – Masacroso
    Nov 8 '19 at 9:40
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    $\begingroup$ I agree with Masacroso. $\sum_{n=1}^{\infty} n = \infty\neq -\frac{1}{12}.$ Don't write such a formula. $\endgroup$ Nov 8 '19 at 9:40
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    $\begingroup$ I did put the word “identity” in quotes. If it were not allowed to ever write down a wrong equation, mathematics would become a few orders of magnitude harder. $\endgroup$
    – celtschk
    Nov 8 '19 at 9:42
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    $\begingroup$ This so-called "identity" was misused so often here that I don't agree with you. There are still many people who just see your first formula and want to believe it. Don't spread fake news. $\endgroup$ Nov 8 '19 at 9:45
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    $\begingroup$ Note that you can write many entire functions $g$ that vanish at $n=1,2,3,...$. Then you can replace $f_n$ with $f_n+\frac{1}{2^n}g$ to get $f+g$ as the sum. Note however, that the question that you asked in the title has another angle besides the one asked at the end. One can ask whether other summation methods give a different answer. Which properties can be kept, etc. $\endgroup$ Nov 8 '19 at 9:53

What if we demand the functions $f_n$ to be analytic rather than just continuous?

No problem. Define

$$f_n(s) = \frac{n}{(n - (-1)^n)^s},$$

where $k^s$ is as usual defined using the real value of $\log k$ (works since $n - (-1)^n > 0$). Then $f_n(0) = n$ for all $n$, and by a standard argument the series converges absolutely and locally uniformly for $\operatorname{Re} s > 2$. We compute \begin{align} \sum_{n = 1}^{\infty} f_n(s) &= \sum_{n = 1}^{\infty} \frac{n}{(n - (-1)^n)^s} \\ &= \sum_{n = 1}^{\infty} \Bigl(\frac{1}{(n - (-1)^n)^{s-1}} + \frac{(-1)^n}{(n - (-1)^n)^s}\Bigr) \\ &= \sum_{n = 1}^{\infty} \frac{1}{(n - (-1)^n)^{s-1}} + \sum_{n = 1}^{\infty} \frac{(-1)^n}{(n - (-1)^n)^s} \\ &= \biggl(\frac{1}{2^{s-1}} + \frac{1}{1^{s-1}} + \frac{1}{4^{s-1}} + \frac{1}{3^{s-1}} + \ldots\biggr) \\ &\qquad + \biggl(-\frac{1}{2^s} + \frac{1}{1^s} - \frac{1}{4^s} + \frac{1}{3^s} - \frac{1}{6^s} + \frac{1}{5^s} - \ldots\biggr) \\ &= \sum_{m = 1}^{\infty} \frac{1}{m^{s-1}} + \sum_{m = 1}^{\infty} \frac{(-1)^{m-1}}{m^s} \\ &= \zeta(s-1) + \eta(s) \end{align} for $\operatorname{Re} s > 2$. This has an analytic continuation to $\mathbb{C}\setminus \{2\}$, and the value at $0$ is $$\zeta(-1) + \eta(0) = -\frac{1}{12} + \frac{1}{2} = \frac{5}{12}.$$

One can by similar means obtain different values.

Such summation methods are however very ad-hoc, as far as I know every "reasonable" summation method assigns either $+\infty$ (the natural value) or $-\frac{1}{12}$ to the divergent series. I admit that I don't know a good definition of "reasonable" for summation methods (except maybe "extends 'limit of partial sums', is linear and stable", but that definition excludes several widely used summation methods).

  • $\begingroup$ How do you get from the sum over $n$ to the sum over $m$? $\endgroup$
    – celtschk
    Nov 8 '19 at 19:09
  • $\begingroup$ Reordering and renaming. I could have also kept the name $n$ of course. The sequence $n - (-1)^n$ goes $2,1,4,3,6,5,\ldots$, so I just swap adjacent terms to get it in the usual order. $\endgroup$ Nov 8 '19 at 19:12
  • $\begingroup$ I see. It would probably make sense to put that information into the post. Nice example! $\endgroup$
    – celtschk
    Nov 8 '19 at 19:19
  • $\begingroup$ @Gerben Where the series converges at all, it converges absolutely, hence it can be reordered in any way we please without changing the sum. The reordering just serves to identify the sum function more easily. $\endgroup$ Nov 13 '19 at 15:21
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    $\begingroup$ @Gerben No. If it is absolutely convergent it can be reordered whether one is interested in the absolute value or not. And we are interested in the function the series defines where it is convergent to see whether that sum function has an (unambiguous) analytic continuation to $0$. Since the series converges absolutely where it converges, reordering is no problem at all. $\endgroup$ Nov 13 '19 at 15:26

Here is a copy of an older question which is more general than yours, but I think, your specific question is included by this:

The initial question MSE

Are there any cases when Abel, Cesaro, Borel, Ramanujan, Zeta regularizations are applicable for regularization of a divergent series or integral but give different results?

Here I cite my own comments (and the interactions of the asker), which may answer your question as far as possible

  • If there were such a case it should be mentioned in wikipedia or mathworld or any serious online/paper encyclopedia ... So I'm sure, there is no such case (except, if you like, that for some regularization the "value" is infinity and for some other is a finite value (multivaluedness in the sense, that only one possible finite value occurs and else only infinities)

  • @Gottfried Helms I need a reference for paper that these approaches are equal.

  • Maybe -but not too modern- Konrad Knopp, "On infinite series" (or so) - it is online in german language, but I think I've come across the english translation as well. Chapter XIII is about divergent series. The other classic is surely G.H. Hardy's monography. But besides of discussion of the Tauberian theorems and perhaps general statements about the equivalence of methods I don't think, they've made the explicite statement in that form you wish here - I surely would remember this! (Note that very similar questions appeared here and in math-overflow from time to time, I think to remember one user who collected material with that focus for writing his batchelor-exposé.)

    Konrad Knopp, unendliche Reihen, chap XIII, pg 480: "(2) permanence-principle: a new regularization-method should first be compatible with convergent series: for convergent series it should give the same value. (2b) Usefulness: But to be useful at all, we expect that it can regularize at least one series which was not convergent itself" (3) When there are different regularization-methods which are capable to assign a finite value to a given series, then all that regularization-methods shall give the same value to that series. " So (3) is -so to say- an axiom which defines, which methods are acceptable in number-theory. Maybe this passage helps for your text (It is not translated here, I just paraphrased. If it is helpful for you you might find it in the english book likely available in google.books - I'm not so good in english to give you a reliable translation by myself)

One should possibly also mention the Springer-online-encyclopedia, which has a lot of entries for divergent summation, and no such ambiguity (two different finite values given by two different regularizations) is mentioned.

  • @Gottfried Helms does he list the methods that give the same value?

  • At least Cesaro, Hölder, Abel, Euler, Riesz,Borel,Le Roy in the book. Don't know at the moment whether he touches Zeta-regularization. Ramanujan is, as far as I remember, in a follow-up article. But I think I've the german chapter locally as pdf-file. I'll see and report S. Chapman & G.H.Hardy (1911) is referred to an article (in "quarterly Journal, Vol 42, pg. 181" on the systematizing of the various regularization-methods. Perhaps there is an explicite statement like you need it.

    Knopp's most contribution is first part of previous century. He did not, for instance know of methods like Aitken-process etc and I do not know whether this compatibility axiom extends to those modern, non-linear, summation procedures as well. Anyway, the reference-list of Knopp's chap XIII should be understandable for you and gives a rich list of further-readings (of course of his contemporaries only...).

another commenter

  • It depends how you define Abel, Cesaro, Borel, Ramanujan, Zeta regularizations and how many of them do you assume well-defined for your series. So make your question precise. The first thing to know is that if an=O(nc) then ?Abel summationan=A well-defined implies F(s)=?ann-s extends analytically to R(s)>0 and lims?0+F(s)=A. math.stackexchange.com/questions/3328567/… @GottfriedHelms

I don't know whether the following would help you as well:

  • I've once considered a type of home-brewn matrix-summation-method using the matrix of Eulerian numbers. In the second part of my explorations I look at the cases, where my method cannot assign a finite value, but at most infinity and somehow a "finite residue" (or so).
    Anyway, instead of getting the Euler-constant $\gamma_0$ for a regularization of the harmonic series as suggested by the Ramanujan-summation-method (or by computation for the second coefficient of the Laurent-series for $\zeta(1+x)$) the value $\log(2)$ pops up - and for the regularization of your series the "residual" value $2/27$ pops up - but I'm not qualified to put this in a reliable and correct context yet. If you like to read then look at my homepage par. 3.3 and 3.4 Perhaps this gives a suggestion and hopefully you can help me to find the correct context.... . *(Surely I'll have to rewrite this, to clear up many messy statements and/or formulations - for instance instead of talking of zeta-series I should talk of polylog-series given my specific framework of matrix-summation - , but perhaps there is still something readable and/or worth for you.)*

The claim $\sum_{n=1}^{\infty} n = - \frac{1}{12}$ depends directly on the Riemann $\zeta$-function. In fact for any constant $c$ one can find a series $\sum a_n(s)$ of functions such that

  • on some subset of the complex plane the series converges and defines a function $f$ that is holomorphic on that subset
  • this function $f$ can be extended meromorphically to the entire complex plane
  • $a_n(-1)=n$
  • the meromorphic continuation of $f$ evaluated at $-1$ is exactly $c$.

If you use $a_n(s)=\frac{1}{n^{s}}$ you get the Riemann $\zeta$-function but that doesn't make $\sum_{n=1}^{\infty} n = - \frac{1}{12}$ any more correct than assigning any other value to this divergent sum.

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    $\begingroup$ It would have been nice if you had given any evidence for your claim. Like a reference, a plausibility argumentation, or a proof sketch. $\endgroup$
    – celtschk
    Nov 8 '19 at 11:58
  • $\begingroup$ @celtschk This is essentially what Daniel Fischer's answer does. But I just upvoted his answer, no reason to spell out the details a second time. $\endgroup$
    – quarague
    Nov 10 '19 at 16:14

Analytic continuation is a very fragile method of summation that often produces inconsistent results depending on what function we choose to continue.

But for this series all the mutually compatible summation methods give the same result.

Moreover, the number $-\frac1{12}$ is just the finite part of the full number. Using approach described here one can obtain the full form of the sum in terms of divergent sums/integrals:

$$\sum_{k=0}^\infty k = \frac{\tau^2}2-\frac1{24}$$

where $\tau=\int_0^\infty dx=\pi\delta(0)$.

Here you can see the full values of the sums of other divergent series: https://extended.fandom.com/wiki/Extended_Wiki#Some_extended_numbers

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    $\begingroup$ What does “mutually compatible” mean in this context? And what about Daniel Fischer's example? It looks very convincing to me, but it seems to contradict what you wrote. $\endgroup$
    – celtschk
    Nov 8 '19 at 19:23
  • $\begingroup$ @celtschk As I said, analytic continuation can bring you just any possible result. $\endgroup$
    – Anixx
    Nov 8 '19 at 19:32

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