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Many textbooks will explain how the Euler product formula can be used as a basis for proving the infinite of primes:

$$\sum_{n}\frac{1}{n^{s}}=\prod_{p}\left(1-\frac{1}{p^{s}}\right)^{-1}$$

My concern is that this is a circular argument because the derivation of the product formula uses the fact there are infinite primes.

Maths history texts will suggest that Euler's derivation of the product formula was done by a process of "sieving the zeta function", outlined next:


We start with the zeta function, having proven (elsewhere) $s>1$ for convergence.

$$\zeta(s)=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\frac{1}{5^{s}}+\frac{1}{6^{s}}+\ldots$$

We can divide this series by $2^{s}$.

$$\frac{1}{2^{s}}\zeta(s)=\frac{1}{2^{s}}+\frac{1}{4^{s}}+\frac{1}{6^{s}}+\frac{1}{8^{s}}+\frac{1}{10^{s}}+\frac{1}{12^{s}}\ldots$$

These denominators are multiples of $2^{s}$. By subtracting these terms from $\zeta(s)$, we sieve out terms with these multiples of $2^{s}$.

$$(1-\frac{1}{2^{s}})\cdot\zeta(s)=1+\frac{1}{3^{s}}+\frac{1}{5^{s}}+\frac{1}{7^{s}}+\frac{1}{9^{s}}+\frac{1}{11^{s}}+\ldots$$

Repeating for $3^s$ gives us:

$$(1-\frac{1}{3^{s}})\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=1+\frac{1}{5^{s}}+\frac{1}{7^{s}}+\frac{1}{11^{s}}+\frac{1}{13^{s}}+\ldots$$

And again for $5^s$:

$$(1-\frac{1}{5^{s}})\cdot(1-\frac{1}{3^{s}})\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=1+\frac{1}{7^{s}}+\frac{1}{11^{s}}+\frac{1}{13^{s}}+\ldots$$

Repeating the process, we can only remove multiples of successive primes. We can't remove multiple of a number C if it is non-prime, C=AB, because we would have already sieved them when we removed multiples of its factors A or B.

In the limit of this process (having removed multiples of an infinitude of primes), only 1 remains on the RHS, and on the left we have an infinite product over all primes:

$$\ldots\cdot(1-\frac{1}{11^{s}})\cdot(1-\frac{1}{7^{s}})(1-\frac{1}{5^{s}})\cdot(1-\frac{1}{3^{s}})\cdot(1-\frac{1}{2^{s}})\cdot\zeta(s)=1$$


Back to my question.

This derivation assumes there are an infinitude of primes in order to attain the single 1 on the RHS. Therefore this Euler product can't be used as basis to say that there is an infinitude of primes.

Where is the flaw in my logic?

Comment: I am aware of the other derivation which uses the FTA and doesn't assume an infinitude of primes, but this is the method used by Euler, and he used it to demonstrate the infinitude of primes, for the first time since Euclid, according to the maths history books.

I am not a university trained mathematician so would appreciate replies with minimal technical terminology.

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    $\begingroup$ In the limit of the process, as you put it, there's only $1$ in the RHS, and all primes in the LHS. No claim about infinitude is made here. It's only when you consider $s=1$ (leaving aside that $\zeta(1)$ is not defined, that's not the point): on the RHS at the top of your derivation, we get the harmonic series, which was known to diverge. Which means that the product over primes (in your last equation) must diverge as well (to $0$), implying there are infinitely many of them. $\endgroup$ Feb 18, 2021 at 14:14
  • $\begingroup$ hi @user3733558 - thanks for trying to help but I'm still not understanding. Here is my thinking: if there were not infinite primes, then the RHS would not be 1. Let's say there are finite primes, eg only {2, 3, 5} .. that would mean the RHS would have a term 1/17^s which can only be removed by having more primes. As long as the number of primes is finite, we will always have terms remaining on the RHS. I would welcome help with my thinking here. $\endgroup$
    – Penelope
    Feb 18, 2021 at 14:47
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    $\begingroup$ I'm not sure I can make it very clear, but it seems to me that you are injecting the assumption of infinite cardinality of the primes into the argument. Euler only says "I've got all of them", regardless of whether there are finitely or infinitely many of them. $\endgroup$ Feb 18, 2021 at 15:07
  • $\begingroup$ Thanks @user3733558 - I will think harder on, what for me, is a subtle difference between "I've got all of them" and the cardinality of the primes. I appreciate your patience. My reason for persisting is that students I teach also find this a difficult point to understand too. $\endgroup$
    – Penelope
    Feb 18, 2021 at 15:11
  • $\begingroup$ Well, it' not easy for me either. I'll try to sum up: Euler's derivation makes no assumption about the cardinality of the primes. The "proof" of that fact is obtained later, by observing that a finite cardinality of the primes would imply convergence of the harmonic series, which is a contradiction. $\endgroup$ Feb 18, 2021 at 15:19

2 Answers 2

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The proofs given below for the Euler product of $\zeta(2)$ and the fact that $\zeta(2)=\frac{\pi^2}{6}$ don't use that there are infinitely many primes.

Now assume that there are only finitely many primes. Then the Euler product for $\zeta(2)$ is a rational number, so that $\frac{\pi^2}{6}$ is a rational number - contradiction.

References:

Euler Product formula for Riemann zeta function proof

Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ (Basel problem)

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  • $\begingroup$ As a curiosity can be mentioned that one of the most classical proofs of that particular value $\zeta(2)$ involve a specific Fourier series expansion which often occurs as one of the first in Fourier Analysis courses. $\endgroup$ Feb 18, 2021 at 14:08
  • $\begingroup$ Yep you are right, appears a bug snuck into my greek letters. $\endgroup$ Feb 18, 2021 at 14:10
  • $\begingroup$ Of course, $s\mapsto 1$ in the Euler product shows that there are infinitely many primes without using that $\pi^2$ is irrational. $\endgroup$ Feb 18, 2021 at 14:32
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You are not using that there are infinitely many primes when expanding the Euler product, just that every integer has a unique factorization in prime powers.

$$\prod_{p\in \text{ set of all primes}} (1+\sum_{k\ge 1}(p^k)^{-s})= \sum_{n\ge 1}n^{-s}$$

For $s > 1$ the LHS is bounded by $\prod_{p\in \text{ set of all primes}} 2$.

That the RHS $\to \infty$ as $s\to 1^+$ implies that there are infinitely many primes.

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  • $\begingroup$ Hi @reuns - in the other common derivation, the use of the FTA is clear. In this derivation, it isn't clear to me where the FTA is used in the "sieving" process. As far as I can see, terms are removed at each step that are multiples of a prime, and these steps are repeated for successive primes. For a number n to be removed it must either be a prime, or a have prime factors. This seems weaker than the FTA. Again, I'm an amateur, so please to tell me where my thinking is wrong. $\endgroup$
    – Penelope
    Feb 18, 2021 at 22:39

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