sequences and series - Strictly speaking, is it true that $zeta(-1)ne1+2+3+cdots$?

There are various notorious proofs that $1+2+3+\cdots=\frac{-1}{12}.$

Some of the more accessible proofs basically seem to involve labelling this series as $S=\sum_\limits{i=1}^ \infty i$ and playing around with it until you can say $12S=-1$.

Even at High School, I could have looked at that and thought "well since you're dealing with infinities and divergent series, those manipulations of $S$ are not valid in the first place so you're really just rearranging falsehoods." It's a bit like the error in this proof that $1=0$, or $\forall x.(\text{false}\implies x)$, it's a collapse of logic.

Greater minds than mine have shown that $\zeta(-1)=\frac{-1}{12}$ and I have no argument with that, but I do dispute the claim that $\zeta(-1)=S$.

My thinking here is that, although the analytic continuation of $\zeta$ is well-defined, that analytic continuation is not the same thing as $\sum_\limits{i=1}^\infty i$.

Once you have

1. defined $\zeta(s) =\sum_\limits{n=1}^\infty\frac{1}{n^s}$ where $\vert s\vert>1$


2. defined $\zeta^\prime(s)=...$ by analytic continuation for all $s$

then you can only claim

3. $\zeta(s)=\zeta^\prime(s)$ where $\vert s\vert>1$.

Basicaly, your nice, differentiable-everywhere definition of the Zeta function is not substituable for the original series $S$ in the unrestricted domain.

Hence, $\zeta(-1)=\frac{-1}{12}\nRightarrow S=\frac{-1}{12}$.

Right? Convince me otherwise.

Answer

You only have

$$\zeta(s)=\sum_{n=1}^\infty n^{-s}$$

for $\mathfrak R(s)>1$. The right-hand side of the equation is not defined otherwise.

Like you said, $\zeta(s)$ is defined by analytic continuation on the rest of the complex numbers, so the formula $\zeta(s)=\sum_{n=1}^\infty n^{-s}$ is not valid on $\mathbb C \setminus \{z\in \mathbb C, \mathfrak R(z)>1\}$.

Therefore,

$$\frac{-1}{12}=\zeta(-1)\ne \sum_{n=1}^\infty n \quad\text{(which $=+\infty$ in the best case scenario)}.$$

So what you say is correct.