TITLE: Intuition behind $\zeta(-1)$ = $\frac{-1}{12}$
QUESTION [9 upvotes]: 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.

REPLY [4 votes]: 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
$$
\begin{align}
\zeta(-2)
&=\lim_{n\to\infty}\left[\sum_{k=1}^nk^2-\frac13n^3-\frac12n^2-\frac16n\right]\\
&=\lim_{n\to\infty}\left[\frac{2n^3+3n^2+n}6-\frac13n^3-\frac12n^2-\frac16n\right]\\[9pt]
&=0\tag{5}
\end{align}
$$