3

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.

Community
  • 1
Jason Smith
  • 1,095
  • 5
    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
  • 2
  • 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
  • 1
    See this and this related question. – J. M. ain't a mathematician 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
  • So, since $\sum 1$ diverges, we take the analytic continuation and then things work out? – Jason Smith Sep 21 '11 at 17:19
  • 1
    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
  • Yet another related question. – J. M. ain't a mathematician Sep 21 '11 at 17:34
  • 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, – Matt E Sep 21 '11 at 18:59
  • This (type of) question was also discussed here: http://math.stackexchange.com/questions/25014/erroneous-numerical-approximations-of-zeta-left-frac12-right/25019#25019 – JavaMan Sep 21 '11 at 20:48