# Why can we extend the zeta function from $\{\Re(s) > 1\}$ to $\{\Re(s) > 0\}$?

First off, let me start by saying that I have not studied complex analysis nor number theory yet, so I may be too eager in my pursuits to study $\zeta(s)$. But I am a quick study, and have found the function mesmerizing since I first saw it's relation to primes in high school.

Why can we extend the zeta function from $\{\Re(s) > 1\}$ to $\{\Re(s) > 0\}$?

As a friendly reminder, $$\zeta(s) = \sum_{n = 1}^\infty \frac1{n^s}$$ which converges for $\Re(s) > 1$ by the integral test, and diverges for all other values.

By multiplying through by $(1-2^{1-s})$ (and later dividing by it), we arrive at an alternate identity:

$$\zeta(s) = \frac1{1-2^{1-s}}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}$$ By the alternating series test, this converges for $\Re(s) > 0$ with $\Re(s) \ne 1$.

In this way, we've extended the domain, however, it seems contradictory to me, as the former definition implies divergence for $0 <\Re(s) <1$, and the later implies convergence in this region.

My thoughts were that the integral test perhaps does not apply the same way for $s \in \mathbb{C}$ as it did for $s \in \mathbb{R}$, but I don't see off-hand why this same dichotomy wouldn't happen if $s \in \mathbb{R}$, as the derivation didn't seem to rely on complex numbers.

Some help understanding this mystery would be greatly appreciated.

• There is a relationship between the Gamma function and the Zeta function that makes this possible. It is discussed in Conway's book on complex variable. – ncmathsadist Jun 5 '17 at 13:51
• How do you apply the alternating series test to a series with complex non-real numbers? – José Carlos Santos Jun 5 '17 at 14:00
• @JoséCarlosSantos: Fair point. As said before, I haven't studied complex analysis, so I was perhaps being too eager again by assuming it applied. – infinitylord Jun 5 '17 at 14:32
• @infinitylord: The alternate sign test simply makes no sense if applied to series of complex non-real numbers because then what do you mean when you say that the terms are alternatively positive and negative? – José Carlos Santos Jun 5 '17 at 14:47
• $\eta(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$ is a convergent alternating series for every $s > 0$ and by the properties of Dirichlet series it implies the convergence and analyticity of this series for every $s \in \mathbb{C}, \Re(s) > 0$. Then you note that for $\Re(s) > 1$ : $\eta(s) = (1-2^{1-s}) \zeta(s)$ thus $\frac{\eta(s)}{1-2^{1-s}}$ has to be the (unique) analytic continuation of $\zeta(s)$ to $\Re(s) > 0, s \not \in 1+ \frac{2i\pi}{\ln 2} \mathbb{Z}$ (where $1-2^{1-s} \ne 0$) – reuns Jun 5 '17 at 22:36

## 1 Answer

The divergence of $\sum_{n = 1}^\infty \frac1{n^s}$ simply means that the extension will not be the limit of this series. It doesn't mean that the limit you get doesn't have a nice extension.

To understand better the phenomena, just look at the geometric series $$\sum_{n=0}^\infty z^n=\frac{1}{1-z}$$

The series is convergent only for $|z|<1$, but this doesn't mean that the function on the RHS only makes sense for $|z|<1$. It only means that outside this disk, the function on the RHS cannot be equal to this particular series.

But this says absolutely nothing about other possible series, and since the function is analytic in the punctured plane $\mathbb C \backslash \{ 1 \}$, there are actually other series equal to this function.

Added Here is an alternate way of thinking about it:

Let $$\zeta(s) = \sum_{n = 1}^\infty \frac1{n^s} \,;\, \Re(s)>1$$ We define a new function $$\zeta_1(s) = \frac1{1-2^{1-s}}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} \,, \Re(s)>0, \Re(s) \neq 1$$

This is a new function, with a larger domain, and we have $$\zeta(s)=\zeta_1(s) \,;\, \Re(s) >1$$

This means that the new function extends $\zeta$, and the theory of Analytic functions tells us that this is the unique extension of $\zeta$ to $\Re(s)>0$. Because of this, we call this new function $\zeta$, even if technically it is a different function.

• I think I understand what you mean, but was a little confused by "the extension will not be the limit of this series. It doesn't mean that the limit you get doesn't have a nice extension." By this, do you mean that the extension I defined will apply for $\Re(s) > 1$, but will not hold equivalence for $0 < \Re(s) < 1$? – infinitylord Jun 5 '17 at 14:29
• If so, is the point of the extension to essentially scrap the original definition and use one that is equal in the original domain and can take values on a larger domain, essentially giving us a more general zeta function? – infinitylord Jun 5 '17 at 14:30
• @infinitylord What I mean is that while $\zeta(s)$ is defined by the new series on $\Re(s) > 0$, the equality $$\zeta(s)=\sum_{n=1}^\infty n^{-s}$$ only holds in the old domain $\Re(s)>1$. – N. S. Jun 5 '17 at 14:35
• @infinitylord Look for "analytic continuation" to understand better this phenomena. It happens often that a power series to be convergent to some function $f$ on some domain. The function $f$ is then analytic on this domain, but often $f$ can be extended to a larger connected domain in the plane. In this case the maximal extension is unique, and agrees with any extension on the domain of the extension. This allow us to think about $f$ as a function defined on a larger domain (its actually its extension, not $f$), but the equality with the power series only holds on the old domain. – N. S. Jun 5 '17 at 14:39
• One reason to expand your interest from the "p-series" perspective to the zeta function you have already expressed: the connection with prime numbers. I assume the connection you saw already was the factorization of the series into a product over primes. But there is a much deeper connection between the distribution of primes and the distribution of "zeroes" of the zeta function, that is, complex numbers which when input produce the output zero. And none of these complex numbers has real part bigger than 1, so you must extend the function to see this connection. – Barry Smith Jun 5 '17 at 14:53