Riemann Zeta as Euler Product

As you likely know, the Riemann Zeta function, commonly expressed as a sum:

$$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$$

can be written as an Euler product formula:

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


For the Riemann Zeta function to be zero, this means at least one of those product factors must be zero:

$$\frac{1}{1-p^{-s}} = 0$$


Why is the above logic considered incorrect? Where is the flaw?

If it is indeed correct, then why don't mathematicians talk about proving the Riemann Hypothesis that non-trivial zeros lie on the $Re(s) = \frac{1}{2}$ line, by focussing on these factors, one of which must be zero?

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    $\begingroup$ The Euler product for the $\zeta$ function and the series $\sum_{n\geq 1}\frac{1}{n^s}$ are converging only in the region $\text{Re}(s)>1$. When we talk about the zeroes of the $\zeta$ function, we mean the zeroes of the analytic continuation of such series. $\endgroup$ – Jack D'Aurizio Apr 21 '17 at 13:42
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    $\begingroup$ And indeed, the Euler product shows that there are no zeros for $\Re(s)>1$. $\endgroup$ – Chappers Apr 21 '17 at 13:43
  • $\begingroup$ And of course, $\frac{1}{1-p^{-s}}$ cannot be zero. $\endgroup$ – Jack D'Aurizio Apr 21 '17 at 13:43
  • $\begingroup$ Thanks Jack and Chappers .. I didn't understand that the definition is only for Re(s)>1. Can you point me to a blog/website explaining why this is the case? $\endgroup$ – Tariq Rashid Apr 21 '17 at 13:49
  • $\begingroup$ And any pointers to learn about "analytic continuation" (for non-professional mathematicians) would be helpful. $\endgroup$ – Tariq Rashid Apr 21 '17 at 13:50

Even ignoring the analytic continuation issue, you've made a fundamental error: treating infinite products in the same way as finite products.

While it is true that if a product of finitely many terms is zero, then one of the terms must be zero, this is false when we consider infinite products. Consider the product $1\cdot {1\over 2}\cdot {1\over 3}\cdot ...$; it's easy to show that this equals zero, even though each term is positive.

Remember that - just like an infinite sum - an infinite product is defined as a limit: $$\prod_{i\in\mathbb{N}} a_i=\lim_{n\rightarrow\infty}\prod_{i\le n} a_i.$$ If each $a_i$ is nonzero, then $\prod_{i\le n}a_i$ is nonzero - but the limit of a sequence of nonzero terms can be zero, so this doesn't stop the whole product from being zero.

(Along the same lines, you might consider why $0.999999...=1$ even though any finite sequence of $9s$ after the decimal point yields a number $<1$.)

  • $\begingroup$ thanks - that's the central point you've answered. $\endgroup$ – Tariq Rashid Apr 21 '17 at 14:26

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