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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?
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    $\begingroup$ 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
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    $\begingroup$ 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. $\endgroup$ – Shakespeare Nov 25 '14 at 10:09
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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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