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When I first watched numberphile's 1+2+3+... = $\frac{-1}{12}$ I thought the sum actually equalled $\frac{-1}{12}$ without really understanding it.

Recently I read some wolframalpha pages and watched some videos and now I understand (I think), that $\frac{-1}{12}$ is just an associative value to the sum of all natural numbers when you analytically continue the riemann-zeta function. 3Blue1Brown's video really helped. What I don't really understand is why it gives the value $\frac{-1}{12}$ specifically. The value $\frac{-1}{12}$ seems arbitrary to me and I don't see any connection to the sum of all natural numbers. Is there any intuition behind why you get $\frac{-1}{12}$ when analytically continue the zeta function at $\zeta(-1)$?

EDIT(just to make my question a little clearer): I'll use an example here. Suppose you somehow didn't know about radians and never associated trig functions like sine to $\pi$ but you knew about maclaurin expansion. By plugging in x=$\pi$ to the series expansion of sine, you would get sine($\pi$) = 0. You might have understood the process in which you get the value 0, the maclaurin expansion, but you wouldn't really know the intuition behind this connection between $\pi$ and trig functions, namely the unit circle, which is essential in almost every branch of number theory.

Back to this question, I understand the analytic continuation of the zeta function and its continued form for $s < 0$ $$\zeta(s)=2^s\pi^{s-1}\sin\frac{\pi s}2\Gamma(1-s)\zeta(1-s)$$ and how when you plug in s = -1, things simplify down to $\frac{-1}{12}$ but I don't see any connection between the fraction and the infinite sum. I'm sure there is a beautiful connection between them, like the one between trig functions and $\pi$, but couldn't find any useful resources on the internet. Hope this clarified things.


marked as duplicate by Dietrich Burde, Rohan, kingW3, Namaste, anomaly Jan 6 '17 at 16:18

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    $\begingroup$ You may actually look for the relation zeta Riemann satisfies in that domain ... that shall clear a lot of things up $\endgroup$ – Tolaso Jan 6 '17 at 12:28
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    $\begingroup$ +1 for escaping the pit of eternal unfortune and beliefs on things like $1+2+3+\dots=-1/12$ $\endgroup$ – Simply Beautiful Art Jan 6 '17 at 12:35
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    $\begingroup$ terrytao.wordpress.com/2010/04/10/… $\endgroup$ – Chappers Jan 6 '17 at 12:37
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    $\begingroup$ The guys at Numberphile should pay £ 1 to whichever decent educational organization in mathematics they want, for every poor soul who arrives on math.se asking about their video... $\endgroup$ – Did Jan 6 '17 at 12:42
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    $\begingroup$ youtube.com/watch?v=jcKRGpMiVTw $\endgroup$ – Did Jan 6 '17 at 12:48

There is a really nice answer by master Terence Tao on his blog.

The Euler-Maclaurin formula, Bernoulli numbers, The zeta function and real variable analytic continuation/

It shows that smoothed sums $\eta$ for $\sum\limits_{n\le N}n^s\,\eta(n/N)$ have a divergent part in $N^{s+1}$ and a convergent part $-\frac{B_{s+1}}{s+1}$.

The second part of the paper shows how it is related to analytics continuation in the complex plane.

  • $\begingroup$ (for $s \in \mathbb{N}$, otherwise there are other terms in $\frac{N^{s+1-k}}{(s+1)\ldots(s+1-k)}$) $\endgroup$ – reuns Jan 6 '17 at 12:47
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    $\begingroup$ @user1952009: Using the Euler-Maclaurin Sum Formula, the first three "divergent terms" are included in my answer. For $s=0$, one of the terms is divergent; for $s=-1$, two of the terms are divergent; and for $s=-2$, three of the terms are divergent. $\endgroup$ – robjohn Jan 6 '17 at 18:07

We have the functional equation for $\;\zeta\;$ :

$$\zeta(s)=2^s\pi^{s-1}\sin\frac{\pi s}2\Gamma(1-s)\zeta(1-s)$$

which allows to extend the usual definition of the zeta function as infinite series to $\;\text{Re}\,s<1\;$, and then:


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    $\begingroup$ If only things like this were more intuitive for the laymen. $\endgroup$ – Simply Beautiful Art Jan 6 '17 at 12:37
  • $\begingroup$ @SimpleArt They are not "intuitive" even for mathematicians...at least not for me. Anyway, when someone messes with the definition domain of zeta function, there doesn't seem to be much options but use the functional equation... $\endgroup$ – DonAntonio Jan 6 '17 at 12:41

An Elementary Non-Proof

Note that $\dfrac{1}{(1-z)^2}=\sum\limits_{k=0}^\infty\,(k+1)\,z^{k}$ leads to $$\beta = 1-2+3-4+\ldots=\frac{1}{\big(1-(-1)\big)^2}=\frac{1}{4}\,.$$ Hence, if $\alpha =1+2+3+\ldots$, then $$\alpha-\beta =4+8+12+\ldots=4\,(1+2+3+\ldots)=4\,\alpha \,.$$ Thus, $$\zeta(-1)=\alpha=-\frac{\beta}{3}=-\frac{1}{12}\,.$$

Hope it helps.


In equation $(10)$ of this answer, it is shown, using the Euler-Maclaurin Sum Formula, that the analytic continuation of the zeta function for $\newcommand{\Re}{\operatorname{Re}}\Re(z)\gt-3$ is given by $$ \zeta(z)=\lim_{n\to\infty}\left[\sum_{k=1}^n{k^{-z}}-\frac1{1-z}n^{1-z}-\frac12n^{-z}+\frac{z}{12}n^{-1-z}\right]\tag{1} $$ Note that for $\Re(z)\gt1$, the terms beyond the sum vanish and we are left with the well-known definition of $\zeta(z)$: $$ \zeta(z)=\sum_{n=1}^\infty n^{-z}\tag{2} $$

For $z=-1$, $(1)$ becomes $$ \begin{align} \zeta(-1) &=\lim_{n\to\infty}\left[\sum_{k=1}^nk-\frac12n^2-\frac12n-\frac1{12}\right]\\ &=\lim_{n\to\infty}\left[\frac{n^2+n}2-\frac12n^2-\frac12n-\frac1{12}\right]\\[3pt] &=-\frac1{12}\tag{3} \end{align} $$

Furthermore, for $z=0$, $(1)$ becomes $$ \begin{align} \zeta(0) &=\lim_{n\to\infty}\left[\sum_{k=1}^n1-n-\frac12+\frac0{12n}\right]\\ &=\lim_{n\to\infty}\left[n-n-\frac12+\frac0{12n}\right]\\[3pt] &=-\frac12\tag{4} \end{align} $$ and for $z=-2$, $(1)$ becomes $$ \begin{align} \zeta(-2) &=\lim_{n\to\infty}\left[\sum_{k=1}^nk^2-\frac13n^3-\frac12n^2-\frac16n\right]\\ &=\lim_{n\to\infty}\left[\frac{2n^3+3n^2+n}6-\frac13n^3-\frac12n^2-\frac16n\right]\\[9pt] &=0\tag{5} \end{align} $$


The following is taken from this answer.

Using the Dirichlet Eta function and integration by parts twice, we get $$ \begin{align} (1-2^{1-z})\zeta(z)\Gamma(z) &=\eta(z)\Gamma(z)\\ &=\int_0^\infty\frac{x^{z-1}}{e^x+1}\,\mathrm{d}x\\ &=\frac1z\int_0^\infty\frac{x^ze^x}{\left(e^x+1\right)^2}\,\mathrm{d}x\\ &=\frac1{z(z+1)}\int_0^\infty\frac{x^{z+1}\left(e^{2x}-e^x\right)}{\left(e^x+1\right)^3}\,\mathrm{d}x\\ \end{align} $$ Multiply by $z(x+1)$ to get $$ (1-2^{1-z})\zeta(z)\Gamma(z+2)=\int_0^\infty\frac{x^{z+1}\left(e^{2x}-e^x\right)}{\left(e^x+1\right)^3}\,\mathrm{d}x $$ Plugging in $z=-1$, gives a pretty simple integral. $$ \begin{align} (1-2^2)\zeta(-1)\Gamma(1) &=\int_0^\infty\frac{e^{2x}-e^x}{(e^x+1)^3}\mathrm{d}x\\ &=\int_1^\infty\frac{u-1}{(u+1)^3}\mathrm{d}u\\ &=\int_1^\infty\left(\frac1{(u+1)^2}-\frac2{(u+1)^3}\right)\mathrm{d}u\\ &=\frac14 \end{align} $$ This gives $$ \bbox[5px,border:2px solid #C0A000]{\zeta(-1)=-\frac1{12}} $$

The relation between $\boldsymbol{\zeta(z)}$ and $\boldsymbol{\eta(z)}$

An alternating sum can be viewed as the sum of the non-alternating terms minus twice the sum of the even terms. $$ \begin{align} \eta(z) &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^z}\\ &=\sum_{n=1}^\infty\frac1{n^z}-2\sum_{n=1}^\infty\frac1{(2n)^z}\\ &=\left(1-2^{1-z}\right)\sum_{n=1}^\infty\frac1{n^z}\\[6pt] &=\left(1-2^{1-z}\right)\zeta(z) \end{align} $$

The integral for $\boldsymbol{\eta(z)\Gamma(z)}$ $$ \begin{align} \int_0^\infty\frac{x^{z-1}}{e^x+1}\,\mathrm{d}x &=\int_0^\infty x^{z-1}\sum_{k=1}^\infty(-1)^{k-1}e^{-kx}\,\mathrm{d}x\\ &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^z}\int_0^\infty x^{z-1}e^{-x}\,\mathrm{d}x\\[6pt] &=\eta(z)\Gamma(z) \end{align} $$



$$\int_{-1}^0x\ dx=-\frac12$$


$$\int_{-1}^0\frac{x(x+1)}2\ dx=-\frac1{12}$$


$$\int_{-1}^0\frac{x(x+1)(2x+1)}6\ dx=0$$

Integrating the formula for the sum of all natural numbers

  • $\begingroup$ I remember seeing this somewhere, but I don't know where... Or why... This works. $\endgroup$ – Simply Beautiful Art Jan 6 '17 at 13:18
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    $\begingroup$ I think this is the closest thing to what I want if only it had an explanation of why/how this works. Anyways, thanks! :) $\endgroup$ – Brian010515 Jan 6 '17 at 13:31
  • $\begingroup$ @Brian010515 I'll try to find or figure out the proof after school, since it'll probably take me a while. $\endgroup$ – Simply Beautiful Art Jan 6 '17 at 13:33
  • $\begingroup$ For whomever, I don't mind downvotes if you explain why you've downoted. Thank you! $\endgroup$ – Simply Beautiful Art Jan 6 '17 at 17:37
  • $\begingroup$ @Brian010515 Please see here. $\endgroup$ – Simply Beautiful Art Aug 23 '17 at 1:03

The values of $\zeta$ for negative integers can be directly calculated from the Bernoulli numbers, from:

$$\zeta(-n)=(-1)^n\frac {B_{n+1}}{n+1}$$

and $B_2=\dfrac 16$.

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    $\begingroup$ I think you've missed the word "intuition" in the question. $\endgroup$ – Wojowu Jan 6 '17 at 12:55
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    $\begingroup$ @Wojowu the question is bizarre. I think that it's impossible to provide an "intuitive" explanation. $\endgroup$ – user384138 Jan 6 '17 at 16:22
  • $\begingroup$ @OpenBall How are you so sure of that? Just because a formal explanation of this is difficult and underlying maths are nontrivial doesn't mean there cannot be a way to explain the result in an intuitive way. $\endgroup$ – Wojowu Jan 6 '17 at 16:24
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    $\begingroup$ @Wojowu I wasn't intending to come across as that sure. I toned my comment down a little bit. I think that there cannot be an intuitive explanation because it is a really conceptually advanced result with much underlying technical aspects. $\endgroup$ – user384138 Jan 6 '17 at 16:27

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