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How come that $sum$ of $all$ positive integers equal a negative rational number.

$$\sum_{n=1}^\infty n = \frac{-1}{12}$$

(original screenshot)

MJD
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AQFarouk
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2 Answers2

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This has been explained on Wikipedia and probably elsewhere on Stackexchange. There are a few ways to understand it, depending on how you understand the series:

  • We are extrapolating the partial sums of the series, which increase with $n$, backward, and it turns out that the correct extrapolation crosses zero to reach a negative number.
  • The leading term of the asymptotic expansion is certainly positive, but that doesn't prevent the constant term from being negative.
  • We are crossing a simple pole in the Riemann zeta function, so we pass from positive values through infinity to negative values. (And we are stopping short of the first trivial zero, so we don't pass back into positive values.)
  • The discretization of the field modes between two conducting plates leads to a positive pressure which is slightly less than the pressure from the modes outside the plates, so the Casimir force ends up being attractive.
  • The second Bernoulli number is positive, and the Euler-Maclaurin formula formally contributes the negative sign. I'm sure these two facts have many more explanations, individually.
Chris Culter
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What is true is that $$ \lim_{r\to-1^+}\sum_{n=1}^\infty nr^{n-1}=\frac14 $$ This limit could be considered a justification for $$ \sum_{n=1}^\infty(-1)^{n-1}n=\frac14 $$ But the erroneous step is to claim that $$ \begin{align} \sum_{n=1}^\infty(-1)^{n-1}n &=\color{#00A000}{\sum_{n=1}^\infty n}-2\color{#C00000}{\sum_{n=1}^\infty 2n}\\ &=-3\sum_{n=1}^\infty n \end{align} $$ because the red sum represents the even terms of the green sum. This manipulation is only valid if the green sum is convergent.

robjohn
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    The last manipulation also turns out to be valid if each sum is (successfully) regularized as a Dirichlet series. So singling out that step as "erroneous" and saying "only valid if ... convergent" is slightly overstating the case. I understand that we don't want the kids at home to abuse similar manipulations where they don't apply. On the other hand, I don't want them to think that rigorous methods can't explain what's going on. – Chris Culter Feb 11 '14 at 03:11
  • @ChrisCulter: perhaps I am missing your point or you are missing mine. Since $\lim\limits_{s\to0^+}\sum\limits_{n=1}^\infty\frac{(-1)^{n-1}}{\Large n^s}=\frac12$, as far as we can take the limit of Dirichlet series which actually converge is $\sum\limits_{n=1}^\infty(-1)^{n-1}=\frac12$. Pushing $s$ to $-1$ takes us outside the neighborhood of an actually convergent series. – robjohn Feb 11 '14 at 10:07
  • We can't get to $s=-1$ with a limit, but we can with analytic continuation, which is part of the definition of zeta function regularization. This is explained in more detail on Wikipedia and on Padilla's website: http://www.nottingham.ac.uk/~ppzap4/response.html – Chris Culter Feb 11 '14 at 10:22
  • @ChrisCulter: yes, I understand the concept of regularization. I am simply pointing out how far we can get without analytic continuation. To make this clearer, I have removed the work regularization from my answer. – robjohn Feb 11 '14 at 10:29
  • Sure, well, it might be better to say precisely that, instead of the absolute characterizations "erroneous" and "only valid if". C'mon, those are fightin' words! :) – Chris Culter Feb 11 '14 at 10:44
  • @ChrisCulter: my point is that the series $\sum\limits_{n=1}^\infty n$ diverges unless we give some alternate meaning to the term convergence. What I said in my answer is true unless the terminology is redefined somehow. This redefinition should be made clear. Using standard definitions, $$\begin{align}\sum_{n=1}^\infty n&=\lim_{n\to\infty}\frac{n^2+n}{2}\&=\infty\end{align}$$ – robjohn Feb 11 '14 at 20:36
  • Yes, the series diverges. I wouldn't want to redefine convergence. The alternating series $1-2+3-4+\cdots$ diverges, yet it is Abel summable, and the Abel sum is defined as a certain limit. The positive series $1+2+3+4+\cdots$ diverges and is not Abel summable, yet it is zeta function regularizable, and the zeta-regularized sum is defined as a certain analytic continuation. These methods are all well-defined; one is not more valid than the other. So I don't see a reason to allow only one of them. – Chris Culter Feb 11 '14 at 21:15
  • @ChrisCulter: I am not saying that we deny the use of regularization. I am simply saying that it would be good to declare that we are using it when we do. If we declare that we are not using normal summation, perhaps so many people won't say $\sum\limits_{n=1}^\infty n=-\frac12$ without qualification. – robjohn Feb 11 '14 at 21:40
  • Agreed, that would be ideal! But then, what does it mean to say "This limit could be considered a justification for ..." in the first step? If we decide that the usual interpretation of the summation symbol must be assumed unless it is explicitly overridden, then that step is also plainly false. – Chris Culter Feb 11 '14 at 22:07