Intuition behind $\zeta(-1)$ = $\frac{-1}{12}$ 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.
 A: 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:
$$\zeta(-1)=\frac1{2\pi^2}\cdot(-1)\cdot1\cdot\frac{\pi^2}6=-\frac1{12}$$
A: 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 $\boldsymbol{\eta(z)\Gamma(z)}$
$$
\begin{align}
\int_0^\infty\frac{x^{z-1}}{e^x+1}\,\mathrm{d}x
&=\int_0^\infty x^{z-1}\sum_{k=1}^\infty(-1)^{k-1}e^{-kx}\,\mathrm{d}x\\
&=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^z}\int_0^\infty x^{z-1}e^{-x}\,\mathrm{d}x\\[6pt]
&=\eta(z)\Gamma(z)
\end{align}
$$
A: 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
$$
\begin{align}
\zeta(-2)
&=\lim_{n\to\infty}\left[\sum_{k=1}^nk^2-\frac13n^3-\frac12n^2-\frac16n\right]\\
&=\lim_{n\to\infty}\left[\frac{2n^3+3n^2+n}6-\frac13n^3-\frac12n^2-\frac16n\right]\\[9pt]
&=0\tag{5}
\end{align}
$$
A: 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. 
A: $$1+1+1+\dots+1=n$$
$$\int_{-1}^0x\ dx=-\frac12$$

$$1+2+3+\dots+n=\frac{n(n+1)}2$$
$$\int_{-1}^0\frac{x(x+1)}2\ dx=-\frac1{12}$$

$$1^2+2^2+3^2+\dots+n^2=\frac{n(n+1)(2n+1)}6$$
$$\int_{-1}^0\frac{x(x+1)(2x+1)}6\ dx=0$$

Integrating the formula for the sum of all natural numbers
A: 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.
A: 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 $B_2=\dfrac 16$.
