Intuition behind $\zeta(-1)$ = $\frac{-1}{12}$

Question

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.

Answer 1

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:

$$\zeta(-1)=\frac1{2\pi^2}\cdot(-1)\cdot1\cdot\frac{\pi^2}6=-\frac1{12}$$

Answer 2

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 $B_2=\dfrac 16$.

Answer 3

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.

Answer 4

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.

Answer 5

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 $$ \bbox[5px,border:2px solid #C0A000]{\zeta(-1)=-\frac1{12}} $$

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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. $$ \begin{align} \eta(z) &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^z}\\ &=\sum_{n=1}^\infty\frac1{n^z}-2\sum_{n=1}^\infty\frac1{(2n)^z}\\ &=\left(1-2^{1-z}\right)\sum_{n=1}^\infty\frac1{n^z}\\[6pt] &=\left(1-2^{1-z}\right)\zeta(z) \end{align} $$

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

Answer 6

$$1+1+1+\dots+1=n$$

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

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$$1+2+3+\dots+n=\frac{n(n+1)}2$$

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

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$$1^2+2^2+3^2+\dots+n^2=\frac{n(n+1)(2n+1)}6$$

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

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Integrating the formula for the sum of all natural numbers

Answer 7

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 $$ \begin{align} \zeta(-2) &=\lim_{n\to\infty}\left[\sum_{k=1}^nk^2-\frac13n^3-\frac12n^2-\frac16n\right]\\ &=\lim_{n\to\infty}\left[\frac{2n^3+3n^2+n}6-\frac13n^3-\frac12n^2-\frac16n\right]\\[9pt] &=0\tag{5} \end{align} $$