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I write $\zeta(s)$ for $\Re(s)>1$ as:

$\zeta(s) = \prod_{p} (1-p^{-s})^{-1}$

Using this I can show that the Riemann zeta function has no zero for $\Re(s)>1$. I'm however not sure about the next step. I would like to use the zero product property, but I know it doesn't necessary hold for infinite products. How do I handle this situation?

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  • $\begingroup$ Take the logarithm (principal branch of the logarithm for each factor). See the series of logarithms converges to a complex number. $\endgroup$ – Daniel Fischer Jan 9 '15 at 17:50
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    $\begingroup$ You want to use Hurwitz's theorem in complex analysis: if an infinite product of nowhere vanishing holomorphic functions on a region in the plane is uniformly convergent on compact subsets then the product function is either nowhere vanishing on that region or is identically zero. Thus the positivity of the Euler product for real $s>1$, or even just at $s=2$, shows $\zeta(s)$ is nowhere vanishing on ${\rm Re}(s) > 1$ if you justify the use of Hurwitz's theorem. $\endgroup$ – KCd Jan 9 '15 at 18:20
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    $\begingroup$ It is a widespread disease in discussions of the zeta-function to say it is nonvanishing on ${\rm Re}(s) > 1$ "because" each Euler factor is nowhere zero on that half-plane. While this can be made rigorous by bringing in Hurwitz's theorem, too frequently that theorem is not mentioned and leads to the mistaken (albeit intuitively appealing) idea that the nonvanishing of the factors alone is all you need, but that is not true. I think the simplest explanation is to write $\zeta(s)$ as an exponential of a holomorphic function on that region, and then use $e^z \not= 0$ for all $z$. $\endgroup$ – KCd Jan 9 '15 at 18:29
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    $\begingroup$ Possible duplicate of Have all the zeros of the Riemann Zeta function real part smaller than 1? $\endgroup$ – A_Sh Jun 14 '17 at 19:52
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The following proposition can be found in 'Complex Analysis' by Stein and Shakarchi (Pg. 141): If $\sum_n |a_n|$ converges then the product

$$ \prod_{n=1}^{\infty}(1+a_n) $$

converges and in this case the product converges to 0 if and only if one of the factors is 0. In this case we have \begin{align*} (1-p^{-s})^{-1}&=\left(\frac{p^s-1}{p^s}\right)^{-1} \\ &=\frac{p^s}{p^s-1} \\ &=1+\frac{1}{p^s-1} \end{align*} so apply the proposition for $a_n=(p^s-1)^{-1}$ to see that the product converges for Re$(s)>1$. Then we know $(1-p^{-s})^{-1}\neq 0$ for all primes $p$ and so using the identity and the second statement in the proposition $\zeta(s)\neq 0$ for all $s\in\mathbb{C}$ with Re$(s)>1$.

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    $\begingroup$ This is all tied up with the definition of the word "converges," which for infinite products does not necessarily mean convergence of the partial products. See my comment to the answer by Markus Schepke, where an infinite product of nonzero factors converges (in the most elementary sense of convergence of partial products) to $0$. $\endgroup$ – KCd Jan 9 '15 at 18:03
  • $\begingroup$ How do you use the given result? What is $a_{n}$ in this case? $\endgroup$ – guestguest Jan 9 '15 at 18:04
  • $\begingroup$ Ah that was a bit confusing of me, I stated the whole proposition however the bit I wanted to emphasize and the only bit I actually used was the fact that if the infinite product converges (which we know it does for Re$(s)>1$) then it is 0 if and only if one of its factors is. $\endgroup$ – S. Mercuri Jan 9 '15 at 18:11
  • $\begingroup$ That is false. Look at the example in my comment to Markus's answer, at $s=1$: $\prod_p (1-1/p) = 0$ and no factor is $0$. As I said before, you have to make it clear what exactly you mean by the term "converges" when you say an infinite product converges. If you just mean the partial products converge then it is wrong that an infinite product of nonzero numbers is nonzero. $\endgroup$ – KCd Jan 9 '15 at 18:16
  • $\begingroup$ Yeah that is true, it is not true in general. The proposition I used states that it is true on the condition given, in your case $\sum_p 1/p$ is not finite so we cannot use this proposition to say anything about it. To apply this proposition to the original question we use $a_n=(p^s-1)^{-1}$, I will edit my answer now. $\endgroup$ – S. Mercuri Jan 9 '15 at 18:26
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It does hold (for absolutely convergent products, see below): just like terms of a sum necessarily need to tend to 0 for the sum to converge, terms of a product need to tend to 1 for it to converge. So you would find zeros in the "finite" factors of the product.

Edit: As pointed out in the comments, you need to be a bit careful with infinite products. However, the statement is true if you consider absolutely convergent products, which is all we need for $\zeta(s)$.

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    $\begingroup$ $\frac{1}{2} \frac{2}{3} \frac{3}{4} \frac{4}{5} \cdots = 0$ $\endgroup$ – 6005 Jan 9 '15 at 17:55
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    $\begingroup$ The issue is not as clear cut as you suggest. The function $1/\zeta(s)$ has the infinite product representation $\prod_{p} (1-1/p^s)$ for ${\rm Re}(s) \geq 1$, including at $s = 1$ where the function vanishes but no factor of the product vanishes. $\endgroup$ – KCd Jan 9 '15 at 18:00
  • $\begingroup$ You're right, I got a little ahead of myself. Now the statement should be more acurate $\endgroup$ – Markus Shepherd Jan 9 '15 at 18:32

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