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

[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

<br /> \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)<br />

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

hmhmh... what kind of set is the set

<br /> \zeta(\{z\in\mathbb{C}\;|\;\textrm{Re}(z)&gt;1\})\subset\mathbb{C}?<br />

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

 

[Hurkyl]

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

 

[jostpuur]

hmhmhm... ok

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

<br /> \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)<br />

 

[Santa1]

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

 

[jostpuur]

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

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

<br /> \prod_{n=1}^{\infty} x_n = 0,<br />

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

 

[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 )

 

[jostpuur]

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

<br /> a_n\to 0<br />

but

<br /> \sum_{k=1}^{n} a_k \to \pm\infty,<br />

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

<br /> e^{a_n} \to 1<br />

but

<br /> \prod_{k=1}^{n} e^{a_k} \to 0\;\textrm{or}\;\infty.<br />

 

[Santa1]

Well that clears that up then .

(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?

 

[Santa1]

As I realize 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}

 

[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)&gt;1 goes like this now. Firstly, we aim for the proof of

<br /> \log |\zeta(z)| &gt; -\infty<br />

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)&gt;1\}). So the task becomes a proof of the claim

<br /> \sum_{p\in\mathbb{P}} \log\big(1 + p^{-2x} - 2p^{-x}\cos(y\log(p))\big) &lt; \infty<br />

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&gt;P with some P. When z is fixed,

<br /> \lim_{p\to\infty}\big( p^{-2x} - 2p^{-x}\cos(y\log(p)) \big) = 0<br />

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

<br /> \underset{p&gt;P}{\sum_{p\in\mathbb{P}}} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \big(<br /> p^{-2x} - 2p^{-x}\cos(y\log(p))\big)^n<br />

converges.

<br /> \big| p^{-2x} - 2p^{-x}\cos(y\log(p))\big| \leq 3 p^{-x}<br />

so actually it suffices to prove

<br /> \underset{p&gt;P}{\sum_{p\in\mathbb{P}}} \sum_{n=1}^{\infty} \frac{1}{n} p^{-xn} &lt; \infty.<br />

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

<br /> \underset{p&gt;P}{\sum_{p\in\mathbb{P}}} p^{-xn} \leq \int\limits_{P}^{\infty} p^{-xn} dp = \frac{P^{1-xn}}{xn-1}<br />

<br /> \sum_{n=1}^{\infty} \underset{p&gt;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} &lt; \infty<br />

 

[Santa1]

Here is a real analysis based proof;

http://mathrefresher.blogspot.com/2006/11/absolute-convergence-for-infinite.html

Similarily here is a real analysis based proof in the case of the zeta function;

http://fermatslasttheorem.blogspot.com/2006/11/euler-product-formula-converges-for.html

And a complex one here;

http://www.math.utah.edu/~taylor/8_InfiniteProducts.pdf