To sum $1+2+3+\cdots$ to $-\frac1{12}$ $$\sum_{n=1}^\infty\frac1{n^s}$$
only converges to $\zeta(s)$ if $\text{Re}(s)>1$.
Why should analytically continuing to $\zeta(-1)$ give the right answer?

Today I stumbled on a totally elementary derivation of this that does not rely on regulators or any reference to the Riemann Zeta Function. We start by looking at the geometric series
$$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + ... $$
Now instead of computing the taylor series of this function we can to compute its $\log$-taylor series that is the coefficients in terms of $\log(x)^n $
First we recall that  $x^n = e^{n \log(x)} $ then it follows that:
$$ \frac{1}{1-x} = 1 + e^{2 \log(x)} + e^{3 \log(x)} +e^{4 \log(x)} + ... $$
Therefore:
$$ \frac{1}{1-x} = 1 + \left(1 + 2 \log(x) + \frac{1}{2!} \left( 2 \log(x) \right)^2+ ...    \right) + \left( 1 + 3 \log(x) + \frac{1}{3!} (3 \log(x))^2 \right) + ...   $$
Which we can re-arrange as:
$$ \frac{1}{1-x} = 1 + \left(1+1+1...  \right) + (1+2+3+4....) \log(x) + \frac{1}{2!} (1^2+2^2 +3^2 ...) \log(x)^2 +... $$
Which obviously don't converge.
But we have ONE more observation. We can get the log taylor series one other way. First we find the laurent series (since theres a negative power term) of
$$ \frac{1}{1-e^x} = -\frac{1}{x} - \frac{1}{2} - \frac{1}{12}x + \frac{1}{720}x^3 + ... $$
So we conclude then that
$$ \frac{1}{1-x} = -\frac{1}{\ln(x)} + \frac{1}{2} - \frac{1}{12}\ln(x) + \frac{1}{720}\ln(x)^3 + ... $$
So then in order for this to be consistent with our earlier result it must be the case that
$$ 1 + \left( 1 + 1 + 1 ... \right) = 0! *\left(  \frac{1}{2} \right) \rightarrow \left( 1 + 1 + 1 ... \right) =  -\frac{1}{2} $$
$$ 1 + 2 + 3 + 4 ... = (1!) *\left( - \frac{1}{12} \right) = - \frac{1}{12} $$
$$ 1^2 + 2^2 + 3^2 + 4^2 ... = 2! * 0 = 0  $$
$$ 1^3 + 2^3 +3^3 + 4^3 ... = 3!* \frac{1}{720} = \frac{1}{120} $$
etc..
CAVEAT:
There is that mysterious $- \frac{1}{\ln(x)}$ that is not accounted for. I'm not really sure what to do about that, although I have some undeveloped ideas brewing which ill update this with once i have refined them.
Another Elementary Approach:
A function $G(x)$ can be written as a series of the form $a_n f(x)^n$ for your choice of $f(x)$ if we define the operator $ D = \frac{1}{f'(x)} \frac{d}{dx}$ and $c$ is a constant such that $\frac{1}{f'(c)}$ is well defined and $f(c)=a$  then
$$ G(x) = G(c) + D[f](c)(f(x)-a) + \frac{1}{2!}D^2[f](c)(f(x)-a)^2 + \frac{1}{3!}D^3[f](c)(f(x)-a)^3 ... $$
Now if $G(x)$ has a singularity $x=c$ as is the case for $G(x)=\frac{1}{1-x}$ $f(x) = \ln(x)$, $c=1$, then you basically need an analogue of the residue theorem to extract its laurent series w.r.t $f$, i'm working on building that right now, but assuming such a thing can be constructed it gives us a different elementary framework which explains the $-\frac{1}{\ln(x)}$ term, but the mystery of why the first method doesn't catch it remains...
