Why is $\zeta(1)^{-1}=0$?

Question

It is known that the Riemann zeta function $\zeta(s)$ has a simple pole at $s=1$. Then why is it that the reciprocal of $\zeta(1)$ is equal to $0$? Shouldn't it be an undefined value?

My source is Wolfram Alpha.

Answer 1

If you make a Taylor series around $x=1$, you have $$\frac 1{\zeta(x)}=(x-1)-\gamma\, (x-1)^2+O\left((x-1)^3\right)$$

Answer 2

There are at least three possible points of view here.

1. In general, if $f$ has a pole at $p$ then $\lim_{z\to p}\frac1{f(z)}=0$. Hence $1/f$ has a removable singularity at $p$.

Now, in complex analysis if $f$ has a removable singularity at $p$ we often speak as though it simply has no isolated singularity at all. For example, people speak of "the entire function $f(z)=\sin(z)/z$", ignoring the fact that $\sin(0)/0$ is undefined, simply taking $f(0)=1$, where more properly they'd define $f$ in cases $z\ne0$, $z=0$. This is just an instance of that convention (I'd be a little happier with $$\frac1\zeta(1)=0$$than with $$\frac1{\zeta(1)}=0$$from this point of view...)

2. The Riemann sphere $S=\mathbb C\cup\{\infty\}$ is a complex manifold. If $f$ is meromorphic in $V$ one can regard $f$ as a holomorphic map from $V$ to $S$, taking the value $\infty$ at poles.

Now if we define $j:S\to S$ by $$j(z)=\begin{cases}1/z,&z\ne0,\infty, \\\infty,&z=0, \\0,&z=\infty\end{cases}$$then $j$ is literally holomorphic, as a map from a complex manifold to itself.

So if $f$ is meromorphic in $V$ and we regard $f$ instead as a holomorphic map from $V$ to $S$ then $j\circ f:V\to S$ is also holomorphic.

Of course people don't use that notation; when they're thinking of things this way they write just $1/f$ and $1/\infty=0$ for $j\circ f$ and $j(\infty)=0$. If we're taking this point of view then it's literally correct to say $$\frac1{\zeta(1)}=\frac1\infty=0.$$

3. Say $M(V)$ is the class of meromorphic functions in $V$. Note that $M(V)$ is a ring with identity. Instead of thinking of the reciprocal of $f$ as defined pointwise, regard $1/f$ as the inverse in this ring. (Here again writing $\frac1\zeta(1)=0$ seems better than $\frac1{\zeta(1)}=0$..)

Answer 3

We should concentrate on the definition of the function $\zeta(s)$ before examining the "value" $\zeta(1)$. This is done neatly and nicely in Marcus' book "Number Fields", chapter 7. The function is defined by the usual series for $Re(s):=x > 1$, and it is analytic on that half-plane by a a classical lemma on "Dirichlet series" (op. cit., lemma 1). Then it is extended to an a priori meromorphic function on the half-plane $x>0$ by applying the same lemma to $f(s)= 1-\frac 1{2^s}+ \frac 1{3^s} - \frac 1{4^s} + ...$ and defining $\zeta(s)=\frac {f(s)} {1-2^{1-s}}$ . The possible poles come from $1=2^{1-s}$, i.e. for $s=1+\frac {2ik\pi}{log2} , k\in \mathbf Z$ . But for $k\neq 0$, one can show that these cannot occur by introducing another series $g(s)=1+\frac 1{2^s}- \frac 2{3^s} + \frac 1{4^s} + \frac 1{5^s}- \frac 2{6^s}...$ and exploiting $\zeta(s)=\frac {f(s)} {1-2^{1-s}}=\frac {g(s)} {1-3^{1-s}}$ (op. cit.). For $k=0$, $\zeta$ does have a pole at $s=1$ because $f(1)\neq 0$, and it is simple because $0$ is a simple zero of $1-2^{1-s}$. Thus the "value" $\zeta(1)$ does not exist in $\mathbf C$, and your "formula" $\zeta(1)^{-1}=0$ just means that $\zeta(s) \to \infty$ when $s\to 1$. What exists is the residue at $s=1$, which can be computed to be $1$.

NB: Of course one could work on the Riemann sphere as in the answer of Davis C. Ulrich, but then the arithmetical meaning of the residue is lost.What arithmetical meaning ? It appears when replacing $\mathbf Q$ by a number field $K$, and the Riemann $\zeta(s)$ by the Dedekind $\zeta_K(s)$. The residue at $s=1$ of $\zeta_K(s)$ is given by the famous "analytical class number formula" which relates, via the residue, the order of the ideal class group of $K$ (an arithmetical object) and the so called regulator of $K$ (a transcendental object).