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I read from some books of number theory that $$\sum_{n=1}^{\infty}\frac{1}{n^s} = -\frac{1}{12}\text{,when } s=-1.$$

Now is there such a result $$\sum_{n=1}^{\infty}\frac{1}{n^s} = \pi \text{,when } s=1,$$or $$\sum_{n=1}^{\infty}\frac{1}{n^s} = c \pi \text{,when } s=1 \text{,where } c \text{ is a rational number ?}$$

I get a similar result in mathematics by analogue, I suspect the result may have some interpretation in physics.

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Thank you for your editing, Qmechanic –  XL _at_China Aug 22 at 8:28
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Zeta function regularization $\zeta(-1)=-\frac{1}{12}$ is also discussed here and here and links therein, and on Math.SE here. Also related Phys.SE post here. –  Qmechanic Aug 22 at 8:46
    
@Qmechanic,thank you –  XL _at_China Aug 22 at 8:47

2 Answers 2

The true fact is the following. Consider $$\zeta(s) := \sum_{n=1}^{+\infty} \frac{1}{n^s} \quad \mbox{with $s\in \mathbb C$ and } Re \:s >1\:. \tag{1}$$ That function, with the said complex domain, is well defined (the series absolutely and uniformly converges) and is a complex analytic function. As a consequence of a well-known theorem on analytic functions, it is possible to extend $\zeta$ outside its original domain into another complex analytic function with a larger domain. Generally speaking, this extension is locally unique. With this extended definition and domain, (1) does not necessarily hold.

As a matter of fact, it is possible to prove that $\zeta$ admits a unique complex analytic extension on the whole complex plane $\mathbb C$ except the point $s=1$, where there is a singularity (a simple pole) which cannot be eliminated even assuming continuity only (which is a much weaker condition than analyticity).

Summing up, there is a unique complex analytic function $\zeta : \mathbb C \setminus \{1\} \to \mathbb C$ satisfying (1) in the open set $Re\:s >1$, it does not satisfy (1) in the rest of its domain, in particular $\zeta(1)$ cannot be defined.

Identities like $$\sum_{n=1}^{+\infty} \frac{1}{n^s} = -\frac{1}{12}\quad \mbox{if $s=-1$}.$$ do not make sense in any cases, because the series in the LHS does not converge for $s=-1$ (evidently!). They make sense referring to the analytic continuation of the originally defined function $\zeta$. In this precise sense, for instance when computing the determinant of unbounded operators with discrete spectrum, are useful in quantum (field) theory.

ADDENDUM. These properties of $\zeta$ are shared by other similar functions constructed out of the spectrum of some elliptic self-adjoint operator like $A:= -\Delta$, defined on a compact Riemannian manifold: $$\zeta_A(s) := \sum_{\lambda \in \sigma(A)} (m_\lambda \lambda)^{-s}\:.\tag{2}$$ Above $m_\lambda$ is the geometric multiplicity of $\lambda$ which is always finite if $A= -\Delta$ also including perturbations, in compact Riemannian manifolds. Formally speaking, $\det A$ is proportional to the partition function of a QFT admitting the Lorentzian version of $A$ as operator of field equations, and when the Euclidean continuation of Lorentzian Killing time gives rise to compact orbits with period $\beta$ (the inverse temperature). The Killing time is the one used to define the static vacuum and the associated thermal (KMS) states. Formally $$\det A = \prod_{\lambda\in \sigma (A)} m_\lambda \lambda\:.$$ However, that productoria generally diverges. Nevertheless the analytic continuation procedure works. Formally (I omit $m_\lambda$ for the shake of simplicity) $$\zeta_A'(0) = \frac{d}{ds}|_{s=0}\left(\sum_{\lambda \in \sigma(A)} \lambda^{-s}\right) = -\sum_{\lambda \in \sigma(A)} \ln \lambda = -\ln \prod_{\lambda\in \sigma (A)} \lambda\:.$$ Thus one can define $$\det A := e^{-\zeta_A'(0)}\tag{3}\:,$$ provided $\zeta_A'(0)$ exist. In general it exists just in the sense of analytic continuation. I mean that the function in (2) turns out to be well defined and analytic for $Re \:s > c_A$ for some real $c_A$ depending on $A$. Moreover that analytic function can uniquely analytically be extended on the whole $\mathbb C$ except for a discrete set of points defining an extended complex analytic function (actually meromorphic). That set does not include $s=0$. Therefore definitions like (3) are safe, at least in principle.

This method can be generalized to compute more complicated objects like the one-loop (thermal) renormalised stress energy tensor in curved spacetime and, it is possible to prove that the procedure is equivalent to more popular ones like the so-called point-splitting method. (I spent part of my initial career dealing with these interesting topics.)

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Thank you,let's wait for more answer or analysis. –  XL _at_China Aug 22 at 8:39

Valter's answer is completely correct, but I'll just briefly expand on it to address the specific values you ask about. The place to go, really, is the Wikipedia page Particular values of Riemann zeta function, which lists mosts of the values of $\zeta(s)$ (which, as Valter explained, equals $$\zeta(s) := \sum_{n=1}^{+\infty} \frac{1}{n^s}$$ when $\operatorname{Re}(s)>1$) that can be expressed without the use of series, or using simpler ones.

For example, the value $\zeta(2)$ is well known to be $\pi^2/6$, and the other positive, even integers have zeta values which are rational multiples of a power of $\pi$.

On the other hand, the value $s=1$ is rather different, because the zeta function has a pole there. This means that there is no way to make the series $$ \zeta(1) := \sum_{n=1}^{+\infty} \frac{1}{n} $$ mean anything other than $\infty$. This series is of course the harmonic series, which is probably the most famous example of a divergent series, and there are multiple simple proofs of why its value must be infinity.

However, if you ask why we insist that $$1+\frac12+\frac13+\frac14+\cdots$$ is infinite, but we're OK with assigning a finite (and negative) value to $$1+2+3+4+\cdots,$$ then I would say that the second series is just a handy way of expressing something else, and should not really have been used in the first place.

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Thank you for your answer –  XL _at_China Aug 22 at 12:21

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