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

  1. $\zeta(s)=\zeta^\prime(s)$ 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.

  • $\begingroup$ that sounds legit. I agree with u. $\endgroup$ – Jorge Fernández Hidalgo Dec 14 '16 at 21:26
  • 2
    $\begingroup$ =D Thank you so much for being one of the people who look at that and thinks logically! $\endgroup$ – Simply Beautiful Art Dec 14 '16 at 22:00
  • $\begingroup$ Notice that when you say $|s|>1$, you probably meant $\Re(s)>1$. $\endgroup$ – Simply Beautiful Art Dec 14 '16 at 22:02
  • 2
    $\begingroup$ "There are various notorious proofs that $1+2+3+...=\frac{-1}{12}$" Actually the rest of your question explains eloquently why there can be no proof that $1+2+3+...=\frac{-1}{12}$... $\endgroup$ – Did Dec 14 '16 at 22:43

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 $\mathbb C \setminus \{z\in \mathbb C, \mathfrak R(z)>1\}$.


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

So what you say is correct.


As everyone has said, the answer to your question is yes. However, I'd like to make the following point:

If, somehow, I can establish a different form (but still analytic) of the Riemann zeta function by manipulating $\zeta(s)=\sum_{n=1}^\infty n^{-s}$ into something else that happens to make sense for new values of $s$, then by analytic continuation, we can define the original function to be the new one, though the original form still won't make sense for the new values of $s$.

This way, some of the weird manipulations you see done with the divergent series can be made correct, as they hold for the original values of $s$ that made sense. Though it is true that most of the manipulations clearly do not.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.