I have two questions concerning infinite series in the context of the Riemann zeta function.

  1. Given the properties of infinite series, why can't we regroup the terms in $\zeta(0)$ in such a way as to give $\zeta(-1)$? i.e.

$$\zeta(0)=\sum_{n=0}^\infty \frac{1}{n^0}=\sum_{n=0}^\infty 1=1+1+1+\ldots=(1)+(1+1)+(1+1+1)+\ldots=1+2+3+\ldots=\frac{1}{1^{-1}}+\frac{1}{2^{-1}}+\frac{1}{1^{-3}}+\ldots=\sum_{n=0}^\infty n=\sum_{n=0}^\infty \frac{1}{n^{-1}}=\zeta(-1)$$

  1. This one might be a lot simpler to answer: why can we assign a value to $\zeta(-1)=\sum_{n=0}^\infty \frac{1}{n^{-1}}$ when the infinite series on the RHS is clearly divergent, i.e. its $n^{th}$ term is always bigger than its $(n-1)^{th}$ term?
  • 1
    $\begingroup$ The series is only defined where it is convergent, that is, $Re(z)>1$. The rest is a meromorphic continuation. $\endgroup$ – Peter Franek Nov 25 '14 at 10:03
  • 3
    $\begingroup$ I think this may be a basic question but it doesn't deserve the downvote; it's clear and concise and the points that need explaining become obvious. Just my opinion. $\endgroup$ – ShakesBeer Nov 25 '14 at 10:09

In short: in a non-absolutely convergent series you can't do things like reorder and group terms because you may get a different answer. In fact, you can reorder the terms in in the sum $1/1-1/2+1/3-1/4+...$ (which in this case does converge, but not absolutely) to give you any real result you like!!! http://en.wikipedia.org/wiki/Absolute_convergence

$\zeta(-1)$ is something quite different. For numbers with $Re(s) \leq 1$ we don't define $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$, but instead as the function which "smoothly" extends this sum which is well-defined on $Re(s) > 1$ to those numbers with $Re(s) \leq 1$. It is, as was mentioned, a meromorphic continuation.

  • $\begingroup$ Thanks for this answer, it is clear and to the point. However, if $\zeta(-1)$ is defined differently for $Re(s)\le 1$ then why do people still say that adding all positive integers results in $-\frac{1}{12}$. Isn't this statement false then? $\endgroup$ – Klangen Nov 25 '14 at 10:24
  • $\begingroup$ @Pickle I believe it is some kind of popular mathematical joke.. $\endgroup$ – Peter Franek Nov 25 '14 at 10:29
  • $\begingroup$ The people at the University of Nottingham seem to think differently: www.youtube.com/watch?v=w-I6XTVZXww $\endgroup$ – Klangen Nov 25 '14 at 10:38
  • $\begingroup$ Ok, it also has a meaning, you are right. It is famous, but its meaning is not so completely straightforward and you need to be careful when performing simple algebraic operations. $\endgroup$ – Peter Franek Nov 25 '14 at 10:41
  • 1
    $\begingroup$ In that case doesn't somebody need to stop this misinformation from spreading? I mean, this can really confuse people! $\endgroup$ – Klangen Nov 25 '14 at 12:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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