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Possible Duplicate:
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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    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 $s=0$.2011-09-21
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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.2011-09-21
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    See [this](http://math.stackexchange.com/questions/39802) and [this related question](http://math.stackexchange.com/questions/8724).2011-09-21
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    @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 twelfth2011-09-21
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    So, since $\sum 1$ diverges, we take the analytic continuation and then things work out?2011-09-21
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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 defined2011-09-21
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    [Yet another related question.](http://math.stackexchange.com/questions/41938)2011-09-21
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    Dear Jason, Regarding analytic continuation in the context of the $\zeta$-function, the answers to this question may be of some help: http://math.stackexchange.com/questions/8724/riemann-zeta-function-and-analytic-continuation Regards,2011-09-21
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    This (type of) question was also discussed here: http://math.stackexchange.com/questions/25014/erroneous-numerical-approximations-of-zeta-left-frac12-right/25019#250192011-09-21

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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.2011-09-21