3
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ 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 $s=0$. $\endgroup$ – Chris Eagle Sep 21 '11 at 17:04
  • 2
    $\begingroup$ 1. is true for $\hbox{Re} \;s > 1$ only... 2. you will have to learn about "analytic continuation" to answer this. $\endgroup$ – GEdgar Sep 21 '11 at 17:04
  • 1
    $\begingroup$ See this and this related question. $\endgroup$ – J. M. is a poor mathematician Sep 21 '11 at 17:05
  • $\begingroup$ @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 $\endgroup$ – lurscher Sep 21 '11 at 17:18
  • 1
    $\begingroup$ 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 $\endgroup$ – lurscher Sep 21 '11 at 17:22
2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Ok this helps, I guess I need to study analytic continuation. $\endgroup$ – Jason Smith Sep 21 '11 at 17:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.