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Why does $1+2+3+\dots = {-1\over 12}$?

Fermat's Dream by Kato et al. gives the following:

  1. $\zeta(s)=\sum\limits_{n=1}^{\infty}\frac{1}{n^s}$ (the standard Zeta function) provided the sum converges.

  2. $\zeta(0)=-1/2$

Thus, $1+1+1+...=-1/2$ ? How can this possibly be true? I guess I'm under the impression that $\sum 1$ diverges.

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marked as duplicate by J. M., Asaf Karagila, Jonas Teuwen, Rasmus, Chris Eagle Sep 21 '11 at 17:35

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As you say, $\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$ provided the sum converges. This says nothing directly about the value of $\zeta(s)$ when this sum diverges, for example when $s=0$. –  Chris Eagle Sep 21 '11 at 17:04
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1. is true for $\hbox{Re} \;s > 1$ only... 2. you will have to learn about "analytic continuation" to answer this. –  GEdgar Sep 21 '11 at 17:04
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See this and this related question. –  J. M. Sep 21 '11 at 17:05
    
@Chris, go and try to explain that to well renowned physicists as Lubos Motl that still assert that the sum itself is what evaluates to minus one twelfth –  lurscher Sep 21 '11 at 17:18
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However, i would be happy with that assertion if i would be shown evidence that any analytic continuation of that sum needs to be equal to the Riemann Zeta wherever it is well defined –  lurscher Sep 21 '11 at 17:22

1 Answer 1

As GEdgar noted, the zeta function is extended to values for which the series diverges via an analytic continuation.

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Ok this helps, I guess I need to study analytic continuation. –  Jason Smith Sep 21 '11 at 17:21

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