# Non-zero zeta function on plane Re(z)>1

1. Dec 28, 2008

### jostpuur

Riemann says that the zeta function doesn't have zeros on the half plane $\{z\in\mathbb{C}\;|\;\textrm{Re}(z)>1\}$, because the sum

$$\log(\zeta(z)) = \log\Big(\frac{1}{\underset{p\in\mathbb{P}}{\prod}\big(1 - \frac{1}{p^z}\big)}\Big) = -\sum_{p\in\mathbb{P}}\log\big(1 - \frac{1}{p^z}\big)$$

remains finite. Well, why does that sum remain finite? Doesn't look quite obvious to me.

hmhmh... what kind of set is the set

$$\zeta(\{z\in\mathbb{C}\;|\;\textrm{Re}(z)>1\})\subset\mathbb{C}?$$

Does it fit in a domain of some logarithm, or does it wind around the origo?

Last edited: Dec 28, 2008
2. Dec 28, 2008

### Hurkyl

Staff Emeritus
Well, $\log(1-x) \approx -x$ (for small x), so it looks plausible.

3. Dec 28, 2008

### jostpuur

hmhmhm... ok

Possible winding problems can probably be avoided by using a real logarithm.

$$\log(|\zeta(z)|) = -\sum_{p\in\mathbb{P}}\log\big(|1 - p^{-z}|\big) = -\frac{1}{2}\sum_{p\in\mathbb{P}} \log\big(1 + p^{-2x} - 2p^{-x} \cos(y\log(p))\big)$$

Last edited: Dec 28, 2008
4. Dec 28, 2008

### Santa1

Well doesn't the Euler product suffice to show that there are no zeroes in re z > 1?

5. Dec 28, 2008

### jostpuur

It can happen that $x_n\neq 0$ for all $n=1,2,3,\ldots$, but still

$$\prod_{n=1}^{\infty} x_n = 0,$$

so $\zeta(z)\neq 0$ is not clear at least merely because the factors $(1 - p^{-z})^{-1}$ are non-zero.

6. Dec 28, 2008

### Santa1

Yes but $$\lim_{n\to\infty} x_n=1$$ in this case so I think it should hold.

I must however confess I have very small knowledge on the area, and might well be mistaken.

(edit: ugly tex =o )

7. Dec 29, 2008

### jostpuur

If $a_1,a_2,a_3,\ldots$ is a sequence such that

$$a_n\to 0$$

but

$$\sum_{k=1}^{n} a_k \to \pm\infty,$$

then $e^{a_1}, e^{a_2}, e^{a_3},\ldots$ is a sequence such that

$$e^{a_n} \to 1$$

but

$$\prod_{k=1}^{n} e^{a_k} \to 0\;\textrm{or}\;\infty.$$

8. Dec 29, 2008

### Santa1

Well that clears that up then :tongue:.

(edit: a moment of clarity)

in that case,

$$\log \zeta(z) = -\sum_p \log(1-p^{-z}) = \sum_p \sum_{n=1}^\infty \frac{p^{-nz}}{n}$$

using the taylor series for log(1-x), we see that the sum remains finite.

Does that hold?

Last edited: Dec 29, 2008
9. Jan 16, 2009

### Santa1

As I realise now that what I said, although looking plausible does not prove anything.

However I looked into absolute convergence for products and it seems that the product,

$$\prod_i 1+a_i$$ converges absolutely iff $$\sum_i a_i$$ does so.

Apply with $$\prod_p \left( 1+\frac{1}{p^s-1} \right)$$ and you end up with the convergence of $$\sum_p \frac{1}{p^s-1}$$

10. Jan 18, 2009

### jostpuur

I agree that the idea in the post #8 is the one that leads into the proof, but there are some details that have to be worked out. Do you have a proof for the claim in the post #9?

Anyway, my proof of the fact $\zeta(z)\neq 0$ for $\textrm{Re}(z)>1$ goes like this now. Firstly, we aim for the proof of

$$\log |\zeta(z)| > -\infty$$

The real logarithm must be used unless somebody knows with certainty that a complex logarithm exists in the image of $\zeta(\{z\in\mathbb{C}\;|\;\textrm{Re}(z)>1\})$. So the task becomes a proof of the claim

$$\sum_{p\in\mathbb{P}} \log\big(1 + p^{-2x} - 2p^{-x}\cos(y\log(p))\big) < \infty$$

The Taylor series of the logarithm cannot be substituted right a way, because many of the terms inside logarithm may not be in domain of convergence $]0,2[$. In order to prove that the series converges, it suffices to prove that it converges when the sum is restricted to values $p>P$ with some $P$. When $z$ is fixed,

$$\lim_{p\to\infty}\big( p^{-2x} - 2p^{-x}\cos(y\log(p)) \big) = 0$$

so we can choose sufficiently big $P$ so that the terms in the logarithm are inside the Taylor series' domain of convergence, and then the task is to prove that

$$\underset{p>P}{\sum_{p\in\mathbb{P}}} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \big( p^{-2x} - 2p^{-x}\cos(y\log(p))\big)^n$$

converges.

$$\big| p^{-2x} - 2p^{-x}\cos(y\log(p))\big| \leq 3 p^{-x}$$

so actually it suffices to prove

$$\underset{p>P}{\sum_{p\in\mathbb{P}}} \sum_{n=1}^{\infty} \frac{1}{n} p^{-xn} < \infty.$$

This can be done with Fubini's theorem and some approximations.

$$\underset{p>P}{\sum_{p\in\mathbb{P}}} p^{-xn} \leq \int\limits_{P}^{\infty} p^{-xn} dp = \frac{P^{1-xn}}{xn-1}$$

$$\sum_{n=1}^{\infty} \underset{p>P}{\sum_{p\in\mathbb{P}}} \frac{1}{n}p^{-xn} \leq P \sum_{n=1}^{\infty} \frac{1}{n^2} \frac{1}{x - \frac{1}{n}} P^{-xn} < \infty$$

11. Jan 18, 2009