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.

  • that sounds legit. I agree with u. – Jorge Fernández Dec 14 '16 at 21:26
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    =D Thank you so much for being one of the people who look at that and thinks logically! – Simply Beautiful Art Dec 14 '16 at 22:00
  • Notice that when you say $|s|>1$, you probably meant $\Re(s)>1$. – Simply Beautiful Art Dec 14 '16 at 22:02
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    "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 $1+2+3+...=\frac{-1}{12}$... – Did Dec 14 '16 at 22:43
up vote 13 down vote accepted

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.

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.

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