$\displaystyle\sum_{n=1}^\infty \frac{1}{n^s}$ only converges to $\zeta(s)$ if $\text{Re}(s) > 1$.

Why should analytically continuing to $\zeta(-1)$ give the right answer?

Alexander Gruber
  • 26,937
  • 30
  • 121
  • 202
  • 37
    It's not clear (to me) what you mean by "the right answer". Since this was migrated from physics, I suspect that you have in mind the use of this summation in string theory. However, mathematically speaking, there is no "right answer" for summation techniques for divergent series; $-\frac{1}{12}$ is simply the answer that analytical continuation yields. See also http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_·_·_· . – joriki May 18 '11 at 04:40
  • 70
    What does the phrase "right answer" mean? If you *define* zeta(s) in the usual way for s with real part greater than 1, it has an analytic continuation to s = -1 and the value there is -1/12. That does not *mean* the series defining zeta(s) for Re(s) > 1 converges at s = -1 with value -1/12 except in a hand-waving Euler kind of way. This question doesn't really have a mathematical meaning as far as I can tell. – KCd May 18 '11 at 04:42
  • 21
    See: http://math.stackexchange.com/questions/37327/infinity-1-paradox, http://math.stackexchange.com/questions/27526/s-11010010010000-1-9-how-is-that, and http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/ for more information too. – JavaMan May 18 '11 at 04:54
  • 8
    Someone should edit the Wikipedia section that says $1+2+3+... = \frac{-1}{12}$ https://en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values – zerosofthezeta Dec 05 '13 at 05:39
  • 10
    (For newcomers) there are a lot of linked questions now, but most relevant, IMO: i) [on (problems with) finding infinite sums via 'shift & add'](http://math.stackexchange.com/q/37327/), ii) [on analytic continuation](http://math.stackexchange.com/questions/41938/), iii) [on assigning values to divergent series](http://math.stackexchange.com/q/632985/) – Grigory M Jan 12 '14 at 11:52
  • 2
    See also [Infinite series are weird](http://skullsinthestars.com/2010/05/25/infinite-series-are-weird-redux/) blog post – Grigory M Jan 19 '14 at 18:49
  • 1
    Watch this video: http://youtu.be/wt6ngy6pDws – zerosofthezeta Apr 08 '14 at 21:33
  • https://www.youtube.com/watch?v=w-I6XTVZXww – jimjim Jan 20 '15 at 03:31
  • 1
    I would classify the result as an eigenvalue. The summation behaves as if it were this value in specific cases, however the properties, operations and definitions of the summation are not limited to those behaviors. It, in fact, breaks axiom of closure, if we consider each term of the sum as an element in the natural numbers -- since we fail to obtain a result that is an element in the naturals, just as with matrices. There are probably other similarities that could be shown as well, but I'm being fairly lazy today. – JustKevin May 06 '15 at 18:02
  • 2
    You can check out :https://www.youtube.com/watch?v=0Oazb7IWzbA – Isomorphic May 26 '15 at 08:15
  • For those people who agree with $1-1+1-1+\cdots=\frac12$: If this is true then according to you'll $\lim\limits_{x\to\infty} \sin(x)=$ [Average of all possible values of $\sin(x)$]=$\frac1{\sqrt{2}}$? (Because you say that the sum converges to half because it is the average of possible values of it) – Aditya Agarwal Aug 20 '15 at 16:24
  • By average of all possible values I mean all positive, if we take negative also, then $f_{avg}=0$ for $\sin(x)$. – Aditya Agarwal Aug 20 '15 at 16:32
  • 2
    Re the three videos linked to in the comments above, [the one by MrYouMath](https://www.youtube.com/watch?v=wt6ngy6pDws) is to be commended while the first Numberphile video is, for its main part, a lazy exercise of self complacency, which caused enough reactions to motivate its authors to post the other video linked above, as a kind of (not very convincing, if you ask me) damage control operation. – Did May 02 '16 at 17:48
  • @JustKevin The moment we tried to evaluate $S=1+2+3+\dots$, it broke the axiom of logic (and I guess it broke perturbation theory). :'( – Simply Beautiful Art Sep 11 '16 at 01:22
  • @AdityaAgarwal Nah, by the MVT and the fact that $\int_0^N\sin(x)dx$ is bounded, the average value of $\sin$ for positive values is $0$, in the limit sense. – Simply Beautiful Art Sep 11 '16 at 01:28
  • 1
    @Did Agreed. I feel Mathologer's video is also enlightening to the layman in at least the dangers of such layman methods while noting at more acceptable ways. https://www.youtube.com/watch?v=jcKRGpMiVTw – Simply Beautiful Art Sep 11 '16 at 01:30
  • 2
    @SimpleArt Indeed, there is a stark contrast between Mathologer's careful, useful, informative approach and Numberphile's nonsense. And one simply has to love how Mathologer very briefly mentions Numberphile's piece at 1:46-1:55... so delicately put! (See also the first comment below the video.) Thanks for mentioning Mathologer's work. – Did Sep 11 '16 at 10:26
  • Take a look at [Ramanujan summation article at wikipedia](http://en.wikipedia.org/wiki/Ramanujan_summation). – Andreas K. Jul 12 '11 at 23:21
  • Yes, $\zeta(-1)=\frac{-1}{12}$, but the formula $\zeta(-n)=\frac{B_{n+1}}{n+1}\,\,\,n\in\mathbb{N}$ leads to an unndefined answer. FYI, $B_{n+1}$ is the $(n+1)$th Bernoulli number. –  Oct 11 '20 at 17:04
  • For fun, see the new video by J $\pi$, where he shows that $1+2+3+\cdots$ can be evaluated to your favorite positive integer, whatever it is. https://www.youtube.com/watch?v=c_ljelgrv50 – GEdgar Mar 06 '21 at 16:36

15 Answers15


there are many ways to see that your result is the right one. What does the right one mean?

It means that whenever such a sum appears anywhere in physics - I explicitly emphasize that not just in string theory, also in experimentally doable measurements of the Casimir force (between parallel metals resulting from quantized standing electromagnetic waves in between) - and one knows that the result is finite, the only possible finite part of the result that may be consistent with other symmetries of the problem (and that is actually confirmed experimentally whenever it is possible) is equal to $-1/12$.

It's another widespread misconception (see all the incorrect comments right below your question) that the zeta-function regularization is the only way how to calculate the proper value. Let me show a completely different calculation - one that is a homework exercise in Joe Polchinski's "String Theory" textbook.

Exponential regulator method

Add an exponentially decreasing regulator to make the sum convergent - so that the sum becomes $$ S = \sum_{n=1}^{\infty} n e^{-\epsilon n} $$ Note that this is not equivalent to generalizing the sum to the zeta-function. In the zeta-function, the $n$ is the base that is exponentiated to the $s$th power. Here, the regulator has $n$ in the exponent. Obviously, the original sum of natural numbers is obtained in the $\epsilon\to 0$ limit of the formula for $S$. In physics, $\epsilon$ would be viewed as a kind of "minimum distance" that can be resolved.

The sum above may be exactly evaluated and the result is (use Mathematica if you don't want to do it yourself, but you can do it yourself) $$ S = \frac{e^\epsilon}{(e^\epsilon-1)^2} $$ We will only need some Laurent expansion around $\epsilon = 0$. $$ S = \frac{1+\epsilon+\epsilon^2/2 + O(\epsilon^3)}{(\epsilon+\epsilon^2/2+\epsilon^3/6+O(\epsilon^4))^2} $$ We have $$ S = \frac{1}{\epsilon^2} \frac{1+\epsilon+\epsilon^2/2+O(\epsilon^3)}{(1+\epsilon/2+\epsilon^2/6+O(\epsilon^3))^2} $$ You see that the $1/\epsilon^2$ leading divergence survives and the next subleading term cancels. The resulting expansion may be calculated with this Mathematica command
1/epsilon^2 * Series[epsilon^2 Sum[n Exp[-n epsilon], {n, 1, Infinity}], {epsilon, 0, 5}]

and the result is $$ \frac{1}{\epsilon^2} - \frac{1}{12} + \frac{\epsilon^2}{240} + O(\epsilon^4) $$ In the $\epsilon\to 0$ limit we were interested in, the $\epsilon^2/240$ term as well as the smaller ones go to zero and may be erased. The leading divergence $1/\epsilon^2$ may be and must be canceled by a local counterterm - a vacuum energy term. This is true for the Casimir effect in electromagnetism (in this case, the cancelled pole may be interpreted as the sum of the zero-point energies in the case that no metals were bounding the region), zero-point energies in string theory, and everywhere else. The cancellation of the leading divergence is needed for physics to be finite - but one may guarantee that the counterterm won't affect the finite term, $-1/12$, which is the correct result of the sum.

In physics applications, $\epsilon$ would be dimensionful and its different powers are sharply separated and may be treated individually. That's why the local counterterms may eliminate the leading divergence but don't affect the finite part. That's also why you couldn't have used a more complex regulator, like $\exp(-(\epsilon+\epsilon^2)n)$.

There are many other, apparently inequivalent ways to compute the right value of the sum. It is not just the zeta function.

Euler's method

Let me present one more, slightly less modern, method that was used by Leonhard Euler to calculate that the sum of natural numbers is $-1/12$. It's of course a bit more heuristic but his heuristic approach showed that he had a good intuition and the derivation could be turned into a modern physics derivation, too.

We will work with two sums, $$ S = 1+2+3+4+5+\dots, \quad T = 1-2+3-4+5-\dots $$ Extrapolating the geometric and similar sums to the divergent (and, in this case, marginally divergent) domain of values of $x$, the expression $T$ may be summed according to the Taylor expansion $$ \frac{1}{(1+x)^2} = 1 - 2x + 3x^2 -4x^3 + \dots $$ Substitute $x=1$ to see that $T=+1/4$. The value of $S$ is easily calculated now: $$ T = (1+2+3+\dots) - 2\times (2+4+6+\dots) = (1+2+3+\dots) (1 - 4) = -3S$$ so $S=-T/3=-1/12$.

A zeta-function calculation

A somewhat unusual calculation of $\zeta(-1)=-1/12$ of mine may be found in the Pictures of Yellows Roses, a Czech student journal. The website no longer works, although a working snapshot of the original website is still available through the WebArchive (see this link). A 2014 English text with the same evaluation at the end can be found at The Reference Frame.

The comments were in Czech but the equations represent bulk of the language that really matters, so the Czech comments shouldn't be a problem. A new argument (subscript) $s$ is added to the zeta function. The new function is the old zeta function for $s=0$ and for $s=1$, it only differs by one. We Taylor expand around $s=0$ to get to $s=1$ and we find out that only a finite number of terms survives if the main argument $x$ is a non-positive integer. The resulting recursive relations for the zeta function allow us to compute the values of the zeta-function at integers smaller than $1$, and prove that the function vanishes at negative even values of $x$.

Luboš Motl
  • 7,798
  • 1
  • 21
  • 26
  • 167
    The fact that other methods also yield the result $-\frac{1}{12}$ doesn't make any of the comments under the question "incorrect". You *interpreted* "right answer" to mean the one that works in physical applications; that's fine (and I suggested in my comment that it was probably meant that way), but that doesn't make it the right answer in any mathematical sense of the term, and nor does the consistency of several summation methods. I'm not aware of any theory of summation of divergent series that provides a definition of what "the right answer" for a resummation is. – joriki May 18 '11 at 06:04
  • 17
    @Joriki, the right answer -1/12 is a deep mathematical result that is relevant not only in physics but also in any branch of maths that cares about the deep relationships between structures such as sums, functions of complex variables, and many other things. The convergent result is not the right result given the most naive mathematical ways to interpret the sum - as a limit of partial sums but it is surely the result of all the most mathematically profound interpretations of the sum. That's also why Euler who was not quite silly knew that the right sum was $-1/12$ long before quantum physics. – Luboš Motl May 18 '11 at 06:43
  • 83
    @Luboš: It seems we have a difference of opinion about the use of "right" in this context. I don't object to you using "right" in the sense that you're using it; what I object to is labelling other views as "incorrect". To call something "incorrect", you'd need a definition of what makes an answer the right answer. As long as you don't have one, it's a matter of taste that you call the consistent result of several different methods "the right answer" and I don't. – joriki May 18 '11 at 06:47
  • 70
    @Luboš: I too object to your label of comments by others as "shallow". Look at the question: it asks specifically about analytically continuing and writes the value as $\zeta(-1)$. That is what I was responding to in my comment. I am well aware that there are other methods of "deriving" the value -1/12 and I had sketched Euler's technique a while ago on Mathoverflow (see my *2nd* answer to http://mathoverflow.net/questions/13130/historical-question-in-analytic-number-theory). You answered a better question than the one that was asked. It's not a reason to say more focused answers are wrong. – KCd May 18 '11 at 07:59
  • 4
    OK, you conclude $\sum n \to \epsilon^{-2} - 1 / 12$ for $\epsilon \to 0$... that would be taken as $\infty$ anyway. – vonbrand Feb 10 '14 at 02:59
  • 3
    It might be but it is a wrong conclusion. The key property of this $\infty$, which may be subtracted by a counterterm, is that it is independent of all independent variables, so it is formally a scale-invariant constant, and $0$ is the number to be associated with it. The fact that the sum of integers equals $-1/12$ is perhaps the most important fact about mathematics among the laymen - and among mathematicians without breadth and depth. – Luboš Motl Feb 10 '14 at 10:52
  • Euler's method, the exponential regulator method and the zeta function approach also all agree that $1+1+1+\cdots = -1/2$. Are there any physical applications of this formula? – Axel Boldt Mar 06 '14 at 16:36
  • 6
    Sure, it's a completely analogous sum. Just like $1+2+3+\dots$ quantifies the ground state energy in 1+1 dimensions (of a string), $1+1+1+\dots$ quantifies the ground state's charge (of a system of free fermions, for example). It's really the reason why the two degenerate states $|0\rangle$ and $c_0|0\rangle$ have ghost numbers $\pm 1/2$, for example. – Luboš Motl Mar 06 '14 at 19:36
  • 1
    @Luboš: That is very cool. Now you could completely wow me if you also had instances in physics of $0+1+2+3+\cdots=5/12$ or $1+0+1+0+\cdots = 0$. – Axel Boldt Mar 08 '14 at 04:54
  • Dear @AxelBoldt, $0+1+2+3+4+\dots $ is also demonstrably and always equal to $-1/12$, and it may be shown by pretty much the same proof. And $1+1=2$. If all entries except for a finite number are zero, then one adds a finite number of terms which always has uncontroversial rules. – Luboš Motl Aug 02 '14 at 04:01
  • 1
    @Luboš: I was referring to the sum $1+0+1+0+1+0+1+\ldots$ which equals 0 according to [Tao's approach](http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/). Similarly, $0+1+2+3+\ldots=5/12$ is easy to prove with his definitions (just add $1+1+1+1+\ldots=-1/2$ to it). Which definitions are you using? – Axel Boldt Nov 19 '14 at 19:17
  • 4
    @LubošMotl Divergent summation methods often lose shift invariance, so $1+2+3+\dots\ne0+1+2+\dots$ is not surprising. This is part of the reason we marginalize these sums with a term like "divergent", because they *don't* follow all the rules we'd like them to. See https://en.wikipedia.org/wiki/Divergent_series#Properties_of_summation_methods, specifically property #3: stability, which is not honored by this sum. – Mario Carneiro Feb 17 '15 at 16:40
  • Dear @MarioCarneiro, some sums may be hard or even ambiguous but I assure you that both $0+1+2+3+\dots $ and $1+2+3+4+\dots$ are equal to $-1/12$. The sum honors everything that needs to be honored to be certain that $-1/12$ is the only right finite value that may be attributed to it. – Luboš Motl Feb 19 '15 at 07:36
  • 19
    @LubošMotl As @${}$Axel points out, any summability method that gives (1) $1+2+3+\dots=-1/12$ and (2) $1+1+1+\dots=-1/2$ and (3) is additive must also assign $0+1+2+\dots=5/12$ and thus is *not* (4) shift invariant, so holding all beliefs (1,2,3,4) leads to contradiction. Which one are you giving up? – Mario Carneiro Feb 19 '15 at 07:44
  • 4
    What is wrong about your calculation is that you are assigning an incorrect value to this $1+1+1+\dots$. In that sum, one must really keep track from which value of $n$ each term $1$ comes from. So the sum $1+1+1+$ starting at $n=1$ is $-1/2$ but if it starts at $n=0$, the sum is $+1/2$, for example. However, no such ambiguity exists for the analogous value of $\zeta(-1)$. Incidentally, the general sum $(n)+(n+1)+(n+2)+\dots$ is equal to $(n-n^2/2) -1/12$. You may check that it is equal to $-1/12$ both for $n=0$ and $n=1$ and it obeys the consistency checks when removing $k$ initial terms, too – Luboš Motl Feb 19 '15 at 09:03
  • $(n) + (n+1) + (n+2) + \cdots = -n(n-1)/2-1/12$, which is, of course, trivial, based on $1+2+\cdots+n=n(n+1)/2$. – FJDU Mar 01 '15 at 20:58
  • I don't think so - my identity for $(n)+ (n+1)+\dots$ works for an arbitrary fractional $n\in R$, too. The value for $n=1/2$ is as important in superstring theory as the value for $n=0$. – Luboš Motl Mar 01 '15 at 22:06
  • I think it would be better to say that both answers are correct. The people that say it converges have an argument, the people that say it diverges have no alternative. This is probably an even more controversial thing to say, but mathematics is inconsistent, we need to learn to deal with that. – Zach466920 Apr 13 '15 at 23:16
  • 6
    We can achieve anything with appropriate basis of definitions. Clearly, the summation, be it "naive" or not is different fundamentally in comparison to the techniques used here. There is no cause to argue about it if we don't first establish what constitutes a sum of a series. If it's the definition based on limit of partial sums then we're finished. Not even the necessary condition is satisfied for $1+2+\ldots $ to converge. The sum of this particular divergent series being EQUAL to $-1/12$ is definitely incorrect. – AlvinL Sep 28 '16 at 07:05
  • 1
    Euler's method can be made rigorous.. – Simply Beautiful Art Jan 06 '17 at 13:06
  • 9
    There is a gulf in understanding between physicists and mathematicians. Mathematician: define what you mean, and then we can discuss it. Physicist: this answer works in a lot of cases so it's right. – Cheerful Parsnip Dec 29 '17 at 04:42
  • 1
    Please, update or remove the link http://www.kolej.mff.cuni.cz/~lmotm275/RUZE/09/node7.html It's broken. – Ivan Kochurkin Dec 08 '19 at 15:17
  • 1
    @Ivan Kochurkin, https://motls.blogspot.com/2014/01/a-recursive-evaluation-of-zeta-of.html?m=1 – Aleksey Druggist Sep 09 '20 at 16:23
  • I've made a thourough study of the [Casimir effect](https://en.wikipedia.org/wiki/Casimir_effect) but still do't comprehend how [Hendrik Casimir](https://en.wikipedia.org/wiki/Hendrik_Casimir) got away with it. The original article is [On the attraction between two perfectly conducting plates](https://www.dwc.knaw.nl/DL/publications/PU00018547.pdf). And there seems to be [empirical confirmation](http://www.mit.edu/~kardar/research/seminars/Casimir/PRL-Lamoreaux.pdf). But some people say it's caused by van der Waals force instead of Zero Point Energy. How can that result in the same stuff? – Han de Bruijn Aug 13 '21 at 14:20
  • Han, mathematicians used to be much more creative and spiritual than today. The sum of integer really IS equal to minus one over twelve and the opposition feels so strong because people are too down-to-Earth, mechanistic, materialistic, primitive. Otherwise van der Waals and Casimir are the same force in the given condition; van der Waals is just the atomic, microscopic description of the same. When you sum over atoms, you get something that can be extracted from a Fourier analysis of fields, via the sum of integers. – Luboš Motl Aug 15 '21 at 06:59

Here is a variant on Lubos Motl's answer:

Let $S = \sum_{n=1}^{\infty} n$. Then $S - 4 S = \sum_{n = 1}^{\infty} (-1)^{n-1} n.$ We will evaluate this latter expression with a regularization similar to Lubos Motl's.

Namely, consider $$\sum_{n=1}^{\infty} (-1)^{n-1} n t^n = -t \dfrac{d}{dt} \sum_{n=1}^{\infty} (-t)^n = -t \dfrac{d}{dt} \dfrac{1}{1+t} = \dfrac{t}{(1+t)^2}.$$
Letting $t \to 1,$ we find that $-3 S = \dfrac{1}{4}$, and hence that $S = \dfrac{-1}{12}.$

To see the relationship between this approach and Lubos Motl's, note that if we write $t = e^{\epsilon},$ then $t\dfrac{d}{dt} = \dfrac{d}{d\epsilon},$ so in fact the arguments are essentially the same, except that Lubos doesn't perform the initial step of replacing $S$ by $S - 4S$, which means that he has the pole $\dfrac{1}{\epsilon^2}$ which he then subtracts away.

As far as I know, this trick of replacing $\zeta(s)$ by $(1-2^{-s+1})^{-1}\zeta(s)$ is due to Euler, and it is a now standard method for replacing $\zeta(s)$ by a function which carries the same information, but does not have a pole at $s = 1$. The evaluation of $\zeta(s)$ at negative integers by passing to $(1-2^{-s+1})\zeta(s)$ and then performing Abelian regularization as above is also due to Euler, I believe. It is easy to see the Bernoulli numbers appearing in this way, for example.

Of course, taken literally, the series $\sum_{n=1}^{\infty} n$ diverges to $+\infty$, so any attempt to assign it a finite value will involve some form of regularization. Analytic continuation of the $\zeta$-function is one form of regularization, and the Abelian regularization that Lubos Motl and I are making is another. I can't quote a precise theorem to this effect (although maybe others can), but with such a simple expression as $\sum_{n = 1}^{\infty} n,$ I'm reasonably confident that any sensible regularization will necessarily yield the same value of $\dfrac{-1}{12}$. (Lubos Motl makes the same assertion in his answer.)

  • 204,278
  • 154
  • 264
  • 488
Matt E
  • 118,032
  • 11
  • 286
  • 447
  • Thanks, @Matt, for the diplomatic work and interpretations. ;-) I actually added this Euler method but your "translation" helped me to understand that this Euler method *is* actually quite manifestly equivalent to the exponential regulator. – Luboš Motl May 18 '11 at 06:45
  • 13
    @Matt: My first example was flawed, but I think my general point still stands that a satisfactory theory of summation of divergent series would have to explain not only why several seemingly different methods yield the same result, but also why other methods don't. Here's a new example (hopefully less flawed): $S-2S=1+3+5+\dotso$ Now we can subtract $2S$ again, and depending on whether we start substracting the $2$ from the $3$ or the $1$, we get $1+1+\dotso$ or $-1-1-\dotso$ First, these should be different. Also, $1+1+\dotso=\zeta(0)=-1/2$, so that would yield $S-4S=-1/2$, $S=1/6$. – joriki May 18 '11 at 06:57
  • 8
    @Matt: (I'm not saying that this can't be explained, or that the results that do coincide do so by chance; to the contrary, I find it intriguing how many of them coincide; all I'm saying is that the ones that don't coincide are also part of the picture, and a satisfactory theory (one that would in my opinion allow us to speak of "the right answer") would have to have something to say about which of these methods are "valid", which aren't and why.) – joriki May 18 '11 at 07:31
  • Dear @joriki, your "other" method is completely legitimate, too. You just have to avoid simple errors. You never get $\zeta(0)$ by the operations. In reality, $S-2S-2S=-3S$ is $1-2+3-4+\dots$ which is equal to $+1/4$ as shown both in Matt's and my answer. And indeed, that's $-3$ times $-1/12$. When summing the terms, you can't randomly "clump" the terms into pairs and give them wrong interpretations because such a manipulation distorts the values of $n$ from which the individual terms came. – Luboš Motl May 18 '11 at 09:40
  • 46
    @Luboš: Instead of insisting on using terms like "simple errors" "in reality" and "wrong interpretations", it would be more helpful if you could specifically point out in which sense you believe that your calculation of $S-2S-2S$ has more merit than mine. As I emphasized, it seems quite likely to me that there is such a difference; but I was looking for an explanation. How did I "randomly clump" terms in my calculation? In which framework does your combination of the terms appear more systematic than mine? (Not rhetorical questions.) – joriki May 18 '11 at 09:55
  • 15
    @joriki: Dear Joriki, I deleted my comment that was made in reference to your deleted comment. As for your revised example (obtaining $\zeta(0)$), there is a "graded" aspect to $S$, which is being disturbed in your computation. (This is what Lubos is referring to when he writes that you "randomly clump" terms.) In terms of Lubos's answer, this disturbing of the grading is not allowed because of dimensional considerations. Mathematically, all I can think to say right now is that both $\zeta$-regularization and Euler regularization involve the graded aspect of $S$, and preserve it. Regards, – Matt E May 18 '11 at 11:58
  • 1
    @joriki This is exactly why I think divergent series are interesting. I think people sense that the subject has to be founded on more rigorous foundations so they are more comfortable with approaching them. – Max Muller May 19 '11 at 12:53
  • 7
    I don't like the first step of your derivation very much, where you compute $S-4S$. Implicitly, this uses the fact that the sums $1+2+3+\cdots$ and $0+1+0+2+0+3+0+\cdots$ have the same value. While true, this is not immediately obvious. For instance, the sum $1+0+2+0+3+0+\cdots$ has a different value (namely $1/24$). – Axel Boldt Mar 09 '14 at 03:10
  • 1
    @AxelBoldt Perhaps it makes more sense to say that $$(1-2^{1-s})\zeta(s)=\eta(s)$$ where $\eta(s)$ is the Dirichlet eta function. This clearly holds true over $\Re(s)>1$, and extends analytically to other values, with a nifty result that $$\eta(s)=\lim_{x\to-1^+}\sum_{n=1}^\infty\frac{x^{n+1}}{n^s}\forall s\in\mathbb C$$ – Simply Beautiful Art Jul 02 '17 at 23:55

What a great method! I tried a few... $$ \sum_{n = 1}^{\infty} \left(\frac{1}{(2 n)^{s}} - \frac{1}{(2 n - 1)^{s}}\right) = \zeta (s) \Bigl(2^{1 - s} - 1\Bigr) $$ put $s=-1$ to get $\sum_{n = 1}^{\infty} 1 = -\frac{1}{4}$

or $$ \sum_{n = 1}^{\infty} \left(\frac{1}{(2 n + 1)^{s}} - \frac{1}{(2 n)^{s}}\right) = \left(1 - 2^{1-s}\right)\zeta(s) - 1 $$ put $s=-1$ to get $\sum_{n = 1}^{\infty} 1 = -\frac{3}{4}$

  • 96,878
  • 7
  • 95
  • 235
  • 7
    You should read Terence Tao's article on this "oddities". – Pedro Jun 11 '12 at 17:37
  • 11
    For clarity: I'm assuming you mean http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/ (which I found through [JavaMan](http://math.stackexchange.com/users/6491/javaman)'s comment on the question. – MarnixKlooster ReinstateMonica Jan 16 '14 at 20:04
  • 1
    Well this is exactly about [an answer I just posted](http://math.stackexchange.com/questions/1994136/alternative-analytic-continuation-to-zeta-not-giving-frac112-for-sum-of/1994162#1994162) – reuns Nov 01 '16 at 06:36

Computation of $\boldsymbol{\zeta(-1)}$ Using Integration by Parts

Here is a direct computation of $\zeta(-1)$ taken from this answer:

Multiply equation $(1)$ from this answer by $x+1$, then integrate by parts twice, to get $$ \begin{align} (1-2^{1-x})\zeta(x)\Gamma(x+2) &=\int_0^\infty\frac{(x+1)xt^{x-1}}{e^t+1}\mathrm{d}t\\ &=\int_0^\infty\frac{(x+1)t^xe^t}{(e^t+1)^2}\mathrm{d}t\\ &=\int_0^\infty\frac{t^{x+1}(e^{2t}-e^t)}{(e^t+1)^3}\mathrm{d}t\tag{1} \end{align} $$ Now we can plug in $x=-1$ into $(1)$ to get $$ \begin{align} (1-2^2)\zeta(-1)\Gamma(1) &=\int_0^\infty\frac{e^{2t}-e^t}{(e^t+1)^3}\mathrm{d}t\\ &=\int_1^\infty\frac{u-1}{(u+1)^3}\mathrm{d}u\\ &=\int_1^\infty\left(\frac1{(u+1)^2}-\frac2{(u+1)^3}\right)\mathrm{d}u\\ &=\frac14\tag{2} \end{align} $$ Since $(1-2^2)\Gamma(1)=-3$, $(2)$ says that $$ \zeta(-1)=-\frac1{12}\tag{3} $$ Naively plugging $n=-1$ into $$ \zeta(n)=\frac1{1^n}+\frac1{2^n}+\frac1{3^n}+\dots\tag{4} $$ yields $$ -\frac1{12}=1+2+3+4+\dots\tag{5} $$ However, the series on the right of $(5)$ is obviously divergent by the Term Test.

Computation of $\boldsymbol{\zeta(-1)}$ Using Euler-Maclaurin Sum Formula

As is shown in $(10)$ from this answer, for $\mathrm{Re}(s)\lt3$, we have $$ \lim_{n\to\infty}\left(\sum_{k=1}^nk^s-\frac{n^{s+1}}{s+1}-\frac{n^s}2-\frac{s\,n^{s-1}}{12}\right)=\zeta(-s)\tag6 $$ Plugging in $s=1$, we get $$ \begin{align} \zeta(-1) &=\lim_{n\to\infty}\overbrace{\left(\sum_{k=1}^nk-\frac{n^2}2-\frac{n}2-\frac1{12}\right)}^{-\frac1{12}\text{ for all $n$}}\\ &=-\frac1{12}\tag7 \end{align} $$

  • 326,069
  • 34
  • 421
  • 800

[19/3/19] Update - ref. 3b1b - Visualizing the Riemann hypothesis and analytic continuation : https://youtu.be/sD0NjbwqlYw - this piece demonstrates utmost clarity on $\zeta(-1)$ evaluation.

:: An explanation of how this is true and why and where it is useful and significant is summarized in 24 by John Baez (as part of the 19/09/08 Rankin Lectures) and yes, indeed this is weird.

The following is the same Euler's method described by Luboš Motl and Matt E in the other answers.

Let us consider the infinite geometric progression " $1, x, x^2, x^3, x^4, \dots$ "

The sum of this progression can be found by the formula $S_\infty = \large \frac{a}{1-r}$,

$$\implies 1 + x + x^2 + x^3 + x^4 + \dots = \frac{1}{1-x}$$ Differentiating with respect to $x$, $$ \implies 0 + 1 + 2x + 3x^2 + 4x^3 + \dots = \frac{1}{(x-1)^2}$$ Let $x = -1$, (Note: In the LHS, you get the alternating counterpart of the natural numbers series) $$\implies 1-2+3-4+5-6+...= \frac{1}{4} $$

$$ \bbox[5px,border:2px solid green]{\therefore \sum\limits_{i=1}^\infty n(-1)^{n-1} = \frac{1}{4}} $$

Subtracting this from $\sum\limits_{i=1}^\infty n$, $$\sum\limits_{i=1}^\infty n -\sum\limits_{i=1}^\infty n(-1)^{n-1} =4+8+12+...=4\sum\limits_{i=1}^\infty n$$ $$\implies 3\sum\limits_{i=1}^\infty n =-\sum\limits_{i=1}^\infty n(-1)^{n-1}=-1/4$$ $$\bbox[5px,border:2px solid lime]{\therefore \sum\limits_{i=1}^\infty n=-\frac{1}{12}}$$

This intuitive reasoning was made famous in a video by the physicist Phil Plait of Numberphile. It is heavily criticized for being sloppy but it shows a simple method that has fooled a large population of its high school audience (as noted in the comments).
Funfact: Googling -1/12 doesn't give you any results
J. W. Tanner
  • 1
  • 3
  • 35
  • 77
  • 6,444
  • 9
  • 40
  • 67
  • 24
    That is not a proof, since you are manipulating expressions to which no meaning is attached. One *can* attach meaning to (some of) them, but that is a somewhat technical thing, and quite non-obvious. (It may be the case that you do know how to cope with these technical difficulties, and that when you write, say, «$1+2+3+4+\cdots$» you h ave in mind something concrete and meaningful, but in that case you should make it explicit) – Mariano Suárez-Álvarez Jan 18 '14 at 02:37
  • 3
    This proof was featured today in a video posted to Reddit. – Steven Gubkin Jan 18 '14 at 03:31
  • 11
    Maybe not a proof but still a nice and simple way how to fool a random high-school audience. Thumbs up. – Jeyekomon Jan 18 '14 at 17:05
  • Dear Nick, You might note in your answer that this is the same Abelian summation method explained in the answers by Lubos Motl and myself, and that it goes back (at least) to Euler. Regards, – Matt E Jan 19 '14 at 02:09
  • @MattE: I'm so sorry. Now that I re-read your answers, I realize I did the same thing. Should I delete this question and repost when I have a more unique answer? – Nick Jan 19 '14 at 04:54
  • 10
    @BalarkaSen, if you take the trouble of looking at history of edits in this post, you will notice that at the time I made the comment above the answer was of a somewhat different nature... – Mariano Suárez-Álvarez Jan 19 '14 at 06:27
  • @Mariano: It was the same thing. All I did different was find the psuedo-sum of the alternating series and use the sigma notation because it looked more mathematical. – Nick Jan 19 '14 at 06:31
  • 1
    Dear Nick, No, I don't think you should delete the answer, since you've given a nice presentation of it. You could just note that the same derivation was presented in other answers --- this helps people coming to the thread get a feeling for how the different answers are related to one another. Cheers, – Matt E Jan 19 '14 at 14:26
  • The word _it's_ in the final paragraph should be _its._ I can't edit without changing at least 6 characters, but you can. – Brian J. Fink Feb 25 '14 at 20:10
  • `that has fooled a large population of it's high school audience` should be `that has fooled a large population of its high school audience`. `Its` is the genitive (possessive) case pronoun. `It's` is a contraction meaning `it is`. – Brian J. Fink Feb 28 '14 at 02:36
  • Never mind. I figured out how to make the correction myself. – Brian J. Fink Feb 28 '14 at 03:26
  • By the way, the reason why searching in Google (or just about any major search engine) for `-1/12` doesn't return any results is that the hyphen is used to delete search terms! :) – Brian J. Fink Feb 28 '14 at 03:45
  • @BrianJ.Fink: Hehe, thank you for that. The hyphen thing was actually on purpose. To really search for it, you have to use quotes `"-1/12"`. – Nick Feb 28 '14 at 06:38
  • I tried that, and Google gave results for `Matthew 5:1-12`, lol. Then I had to click on the link that said `Show results instead for "-1/12"`. Hee hee – Brian J. Fink Mar 03 '14 at 00:09
  • 4
    You manipulate that infinite sum as if it were finite, like when you assume that $\sum_{n=1}^\infty n-4\sum_{n=1}^\infty n=-3\sum_{n=1}^\infty n.$ But that's impossible, since $\infty-\infty$ is undefined. – Hakim May 16 '14 at 20:52
  • @StevenGubkin, as a redditor you should know that there should always be links. – GinKin Jun 17 '14 at 19:40
  • 2
    $1 + x + x^2 + x^3 + x^4 + \dots = \frac{1}{1-x}$ is true if $-1 –  Jan 05 '15 at 08:43
  • The reason `-1/12` has no Google results is because any phrase starting with `-` has no Google results, as this is a negative operator. – gerrit Jul 26 '16 at 09:34
  • 2
    This can be made more rigorous by instead establishing the relationship between the Riemann zeta function and the Dirichlet eta function when they converge, which gives a functional equation that holds by analytic continuation to $s=-1$. The next step would be to show $\eta(-1)=\frac14$ by showing the method you used above holds analytically. That is, show that $f(s)$ is analytic and equal to the Dirichlet eta function where we define $f(s)$ as $$f(s)=\lim_{x\to-1^+}\sum_{n=1}^\infty\frac{x^{n+1}}{n^s}$$ – Simply Beautiful Art Jun 19 '17 at 00:22
  • @Nick That's Tony Padilla, not Phil Plait. Phil Plait wrote an article about the video, and does not appear in it. – Robert Furber Aug 06 '18 at 20:34

If the following were true: $$\sum_{n=1}^\infty{n}=-\frac1{12}\tag{hypothesis}$$ then we would expect the following: $$\lim_{n\to\infty}\sum_{i=1}^n{i}\\ =\lim_{n\to\infty}\frac{n(n+1)}2=-\frac1{12}\tag{expectation}$$ which is the formula for the infinite triangular number limit. Unfortunately this is a result that we do not get when the limit is correctly taken. The correct value is $$\lim_{n\to\infty}\frac{n(n+1)}{2}\\ =\lim_{n\to\infty}\frac{n^2+n}{2}\\ =\lim_{m:{n^2+n}\to\infty}\frac{m}{2}=\infty\\ \neq-\frac{1}{12}$$ This sort of mathematical sleight of hand, smoke and mirrors, pulling a finite negative rabbit out of an empty positively infinite hat does not impress me; worse yet, it gives legitimate, observable, repeatable mathematics a bad name.

Brian J. Fink
  • 1,330
  • 11
  • 17
  • 17
    Great answer! You should keep the "lim n to infinity" operator in front of every expression until the end instead of simply plugging in infinity... – zerosofthezeta Mar 13 '14 at 03:33
  • 1
    Not sure that is expectable. I made a [question](http://math.stackexchange.com/questions/416536/is-every-sum-to-something-a-limit-of-the-sum) related to it. – JMCF125 Apr 27 '14 at 20:10
  • 8
    I can't tell if this was meant to be a joke, or not. $∞²$ is a nice touch. – primo May 29 '14 at 10:24
  • Sum of all natural numbers=negative fraction.This mathematical truth is too bitter. Isn't it violating the basic rules of mathematics(i.e. sum of positive natural numbers is positive). This mathematical phenomenon or behaviour is weird and incomprehensible. Doesn't this kind of situation tell us that human mind is insufficient to understand everything; may be thought is not capable enough to grasp the nature of every truth. – Sara Tancredi Jun 05 '14 at 23:47
  • @Sara: what about the sequence $1/10,1/9,1/8,1/7,1/6,1/5,1/4,1/3,1/2,1/1,1/0,1/-1,1/-2,...$ After the numbers increased to infinity, then -somehow naturally- by continuation of the process we arrive at negative numbers... and again increasing... I myself do not put too much weight on this example, but it is due to L. Euler - someone who has shown much explanatory (and educational) power, so it might be interesting here as well as a "mind-opener". – Gottfried Helms Jun 06 '14 at 12:45
  • 6
    @GottfriedHelms, did you just divide by 0 ? – GinKin Jun 17 '14 at 19:44
  • @GinKin: no, that was not the focus of my post. I just reproduced an idea of L. Euler how to give an idea, under which circumstances one might change his/her intuition. He (and then me) could have just omitted the "1/0"-expression or have replaced it by a "1/$\delta$"-expression - but that makes the core-idea only worse readable. The sequence in my comment is just a scribble, in no way a mathematical theorem or so, a sketch to make someone think - at least as I've understood Euler with this. – Gottfried Helms Jan 29 '15 at 20:40
  • It is totally possible to generalize the notion of limit. – Anixx Apr 26 '20 at 23:49
  • 1
    @NikhilKumarSingh Check it again. $\frac{n(n+1)}{2}$ is the value of the entire sum, not a single term in it. – Brian J. Fink May 22 '21 at 19:32

This infinite series is ultimately divergent because $$ 1+2+3+4+\cdots=\sum\limits_{k=1}^{\infty} k$$ $$ = \lim\limits_{n\to\infty} \sum\limits_{k=1}^{n} k = \lim\limits_{n\to\infty} \frac{n(n+1)}{2} = \infty $$ The value of $-\frac{1}{12}$ is attained by applying a different summability method, such as zeta function regularization and several others. If you'd like to understand how a divergent series can report a finite value by applying a different summability method, then have a look at this question.

  • 8,551
  • 3
  • 21
  • 43

The notation "$1+2+3+\cdots$" is as meaningless as "$1/0$". If you treat such notation as though it defined a real number and conformed in its syntax to the rules of formation for genuine real numbers, you can easily "prove" it to equal any number you like, including $-1/12$.

John Bentin
  • 16,449
  • 3
  • 39
  • 64
  • 27
    Dear John, Your claim that this notation is "meanlingless" seems unnecessarily absolute, in light of the answers explaining that it does in fact admit meaningful interpretations. Regards, – Matt E Sep 08 '12 at 19:11
  • 10
    Dear Matt: Sure, it admits meaningful interpretations, for some people. But they are not consistent. – John Bentin Sep 09 '12 at 12:03
  • 6
    The fact that so many different methods which connect to deeper waters in mathematics all lead to the same sum suggest that, even though there is currently no way to make it precise, there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet. To me, these are some of the most exciting things in mathematics: having only a glimpse of something great just beyond the horizon. – Steven Gubkin Jan 18 '14 at 03:29
  • 14
    @StevenGubkin Replace "there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet" by "there exists a rigorous consistent theory of divergent sums, made fully precise more than 80 years ago". As Littlewood explains in the preface of Hardy's treatise, "[I]n the early years of the century the subject [Divergent Series], while in no way mystical or unrigorous, was regarded as sensational, and about the present title, now colourless, there hung an aroma of paradox and audacity"... only these were the first years of the 20th century, not the 21st. – Did Jan 18 '14 at 21:50
  • @Did: I understand that we have a partial theory of summability of divergent series. I do not think that we have quite heard the last word on the story. I think the most exciting developments are still to come! In particular the whole "seven trees in one" story seems to leave a lot to be explored. – Steven Gubkin Jan 19 '14 at 15:06
  • 1
    I think everything said in this post is true... but none of it explains _why_ $-1/12$ is a better choice of "sum" than any other real number. (If you think $-1/12$ is no better than any other real number here, well, just read the other answers.) – Jesse Madnick Jan 22 '14 at 01:36
  • 1
    @JesseMadnick: Well, "better" is a matter of taste. If you prefer to involve $\zeta(0)$ rather than $\zeta(-1),$ as in joriki's comment to Matt E's answer, you get a different result. But there is no need to involve the Riemann zeta function. More elementary monkey business can derive whatever result you think best. – John Bentin Jan 22 '14 at 09:16
  • 1
    @JohnBentin: Sure, but the question asked by the OP is "Why $-1/12$?" – Jesse Madnick Jan 22 '14 at 14:00
  • @StevenGubkin You may be interested in this post: https://mathoverflow.net/questions/115743/an-algebra-of-integrals/342651#342651 – Anixx Apr 26 '20 at 23:53

I am missing here that

$$1+2+3+\cdots \rightarrow \infty$$

$$\zeta(-1) \neq 1+2+3+\cdots$$

As you say, we only define $\zeta$ using the infinite sum if $\Re(s)>1$.

It is just as defining $$f(x) = \begin{cases} \frac{1}{x} &\mathrm{if} \; x \neq 0 \\ 0 & \mathrm{if} \; x = 0 \end{cases}$$

And asking why $f(0)=0$.

  • 24,314
  • 6
  • 55
  • 112
  • 1
    your example is not correct. In first place the zeta function defined at values where $\Re(z)\le1$ is defined by analytic continuation. However your function $f$ is not defined by analytic continuation. – Masacroso Jan 17 '18 at 11:34

A nice alternative representation, by a double sum.

Consider the infinite square array where the rowsums and the (formal) column-sums are also given in closed forms $$ \small \begin{array} {r|rrrrr|rr} & & & & & & \text{rowsums} \\ \hline & 1/0! & \log(1)/1! & \log(1)^2/2! &\log(1)^3/3! & \cdots & = e^{\log(1)}&=1 \\ & 1/0! & \log(2)/1! & \log(2)^2/2! &\log(2)^3/3! & \cdots & = e^{\log(2)}&=2 \\ & 1/0! & \log(3)/1! & \log(3)^2/2! &\log(3)^3/3! & \cdots & = e^{\log(3)}&=3 \\ & \vdots & & & & \ddots & \vdots & \vdots\\ \hline \text{colsums:} & \zeta(0) & -\zeta(0)'/1! & \zeta(0)''/2!& -\zeta(0)^{(3)}/3! & \cdots &&=\zeta(-1) \end{array} $$ and the row-sum of the derivatives of the $\zeta()$ at $0$ in the bottom row can numerically be written as $$ \small -0.5 +0.9189... -1.003178...+1.000785...-0.999879...+1.0000019... \pm \cdots $$ If we split the terms and do two series we find $$ \Tiny \begin{array} {} -1 &+1 &-1 &+1 &-1 &+1 &\pm &\cdots &=-1/2\\ +0.5 &-0.0810... &-0.003178...&+0.000785...&+0.000120...&+0.0000019... &\pm &\cdots &= 5/12\\ \end{array}$$ Here we must now resort to the earlier definition of the divergent sum of the alternating units (in the first row), but the second row is convergent and can conventionally be summed. The sum of the two rowsums $\small -1/2 + 5/12 = -1/12$ gives the expected result.

However, that fiddling with double-sums, when non-convergent-series are involved (as are the columnsums in the above matrix and the splitting in the numerical expression) must explicitely be justified as "legal"/algebraically consistent operation.
But because I encounter it not frequently, that such non-convergent double-sum schemes come out with the expected result without further ado, I think this is a specific nice observation here.

And the more algebraically manipulations come out to be consistent with the assumed value of a divergent series, the more is the hypothese acceptable, that this should be taken as the canonical numerical replacement for the series-expression ( like we do it with the rational fraction for the geometric series even in the divergent case (except for that with quotient $q=1$)) .

Gottfried Helms
  • 32,738
  • 3
  • 60
  • 134

Suppose a rigorous way of doing a computation yields a well defined real number as the answer. But with shortcut formal manipulations that are not allowed (e.g. interchanging summations and integrations when that isn't allowed) one ends up with a divergent series and then the question is that given only the divergent series, can one guess what real number is most likely the answer to what the unknown original problem was.

This is similar to guessing that the next term of the integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 is probably 16 rather than e.g. 32431, even though the latter possibility cannot be ruled out. Strictly speaking this problem is not well defined, what one really is doing here is assuming that the sequence that is specified by the least amount of information is the most likely answer. The person who invented the puzzle had some simple algorithm in mind, therefore the much more complicated algorithm that would yield 32431 as the next number in the sequence is not the likely answer.

Similarly, the formal manipulations that yield the divergent series won't have introduced a lot of additional information, these are just generic mathematical manipulations that would have been correct when used in a wide class of problems, but not the one it actually has been applied to. This means that almost all the information about the unknown real number is present in the divergent series, it can be extracted from it by applying certain formal manipulations to the series that, like the unknown manipulations that led to the divergent series, are formally correct for a wide class of convergent series, but not in case of this divergent series.

Count Iblis
  • 10,078
  • 2
  • 20
  • 43

Euler's approach, but using algebra instead of subtraction of infinite series: Let $f(x)=(1-x)^{-2}$. Then $f(-x) = f(x) -4xf(x^2)$. Plug $x=1$ to get $f(-1)=-3f(1)$, whence $f(1)=-1/12$.

EDIT: This argument is a motivation for the mysterious value $-1/12$. Yes, $f(x) :=(1-x)^{-2}$ is undefined at $x=1$. If we define $f(1)=-1/12$, then the above identity holds for all $x$.

  • 32,010
  • 1
  • 28
  • 54

I like this simple graphical explanation.

There are first partial sums of the series 1 + 2 + 3 + 4 + ⋯ on picture. The parabola is their smoothed asymptote; its y-intercept is −1/12.

enter image description here

Ivan Kochurkin
  • 1,075
  • 1
  • 11
  • 25
  • 8
    I'm afraid I don't follow this at all. – Cheerful Parsnip Sep 11 '15 at 21:02
  • @GrumpyParsnip It's explained at https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF#Cutoff_regularization . I'm interested in where the version with the labeled axes came from, though. KvanTTT, did you add those yourself? – Chris Culter Sep 13 '15 at 21:01
  • @ChrisCulter, yep, it's my additions. I updated the answer. – Ivan Kochurkin Sep 27 '15 at 18:12
  • How is this smoothing be done? Excel gives me a trendline (polynomial order 2) which passes exactly through zero. – Gottfried Helms Nov 27 '15 at 20:05
  • @GottfriedHelms, I don't know. Probably there is a way to evaluate not only _c_ coefficient (-1/12) of this parabola, but _a_ and _b_ too. It works only with infinite number of points (excel can process only finite set). – Ivan Kochurkin Nov 27 '15 at 23:40
  • Ah, perhaps it is the integral-formula (which leads then to the constant at zero). Ramanujan-summation works with such an integral for the constant term... something like that... – Gottfried Helms Nov 28 '15 at 06:26
  • 7
    It is somewhat odd that this completely opaque "answer", which the OP themselves concedes they do not know what it means or where it comes from, received 9 upvotes. Maybe for the *colors* in the picture? – Did Dec 04 '16 at 09:42
  • The graph of $y=x^2-\frac18$, which has $y$-axis intercept $-\frac18$, cuts each stage in the middle. If you raise the graph a little bit (by $\frac1{24}$), you can nudge the $y$-axis intercept to $-\frac1{12}$ ; but then the graph intercepts the stages at different (off-centre) places. In your diagram, this is visible on the zeroth stage, where the intercept is at $x=1/\surd6$, and just noticeable on the first stage. – John Bentin Oct 12 '19 at 07:34

As I recently showed in another answer, we have the wonderful pattern:

$$\sum_{k=1}^n1=n\implies\int_{-1}^0x~\mathrm dx=\zeta(0)\\\sum_{k=1}^nk=\frac{n(n+1)}2\implies\int_{-1}^0\frac{x(x+1)}2~\mathrm dx=\zeta(-1)\\\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6\implies\int_{-1}^0\frac{x(x+1)(2x+1)}6~\mathrm dx=\zeta(-2)\\\sum_{k=1}^nk^3=\left[\frac{n(n+1)}2\right]^2\implies\int_{-1}^0\left[\frac{x(x+1)}2\right]^2~\mathrm dx=\zeta(-3)\\\vdots$$

This pattern works for all $\zeta(-k)$.

Simply Beautiful Art
  • 71,916
  • 11
  • 112
  • 250

Here is a useful place for Euler's transform and the Dirichlet eta function:




$$\zeta(s)-\eta(s)=2^{1-s}\zeta(s)\implies (1-2^{1-s})\zeta(s)=\eta(s)$$


After a quick application of Euler's transform, we get a nice analytic continuation to the entire complex plane.

$${\small E_1}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^s}=\sum_{n=0}^\infty\left[\frac1{2^{n+1}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{(k+1)^s}\right]$$

Finally giving us

$$\zeta(s)=\frac1{1-2^{1-s}}\sum_{n=0}^\infty\left[\frac1{2^{n+1}}\sum_{k=0}^n\binom nk\frac{(-1)^k}{(k+1)^s}\right]$$

And at $s=-1$,


John Bentin
  • 16,449
  • 3
  • 39
  • 64
Simply Beautiful Art
  • 71,916
  • 11
  • 112
  • 250