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 $s &lt; 0$ $$\zeta(s)=2^s\pi^{s-1}\sin\frac{\pi s}2\Gamma(1-s)\zeta(1-s)$$ 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.
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.
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 $B_2=\dfrac 16$.