The following Euler product for the Riemann zeta function is well known.

$$ \sum_n \frac{1}{n^s} = \prod_p (1-\frac{1}{p^s})^{-1} $$

Here $n$ ranges over all integers, $p$ over a primes, and real $s>1$.

Common Proof Strategy

Many derivations / proofs, found in textbooks and papers, consider the following expression.

$$(1 - \frac{1}{p^s})^{-1} = 1 + \frac{1}{p^s} + \frac{1}{p^{2s}} + \frac{1}{p^{3s}} + \ldots$$

The LHS is finite for any given $p$ and the series expansion is valid because $\frac{1}{p} < 1$.

The following takes the product over all primes.

$$\prod_{p_i} (1-\frac{1}{p_i^s})^{-1} = 1 + \frac{1}{p_1^s} + \frac{1}{p_1^{2s}} + \ldots + \frac{1}{p_1^sp_2^{s}} + \frac{1}{p_1^sp_3^{s}} + \ldots $$

The LHS is a product of finite and non-zero factors.

The RHS has terms of the form $\frac{1}{X}$ where $X$ contains all combinations of the primes, and all combinations of powers of the primes.

It is common to apply the Fundamental Theorem of Arithmetic to see that there is one term X for each integer $n$, and therefore the RHS is the desired $\sum\frac{1}{n^s}$.


A challenge (for example here) that has been raised to this very common proof logic is that there are terms $X$ with an infinite number of factors in the denominator, for example:

$$ \frac{1}{(2^2\cdot3^2\cdot 5^2\cdot 7^2 \cdot\ldots)^s} $$

or another simpler example:

$$ \frac{1}{(2\cdot 2 \cdot 2\cdot 2\cdot\ldots)^s} $$


Is the challenge valid?

I am not a trained mathematician, but in my opinion the proof strategy is valid because terms $X$ with denominators with an infinite number of prime factors are equivalent to zero. That is:

$$ \frac{1}{(2^2\cdot3^2\cdot 5^2\cdot 7^2 \cdot\ldots)^s} = 0$$


$$ \frac{1}{(2\cdot 2 \cdot 2\cdot 2\cdot\ldots)^s} = 0$$

My assertion is that the proof strategy remains valid because any finite integer $n$ has a single finite non-zero term $X$, and those $X$ with infinitely long denominators can be discarded because they are zero.

  • 5
    $\begingroup$ There aren't really any terms with an infinite product in the denominator. The infinite product is really a limit of finite products, and the terms with "infinitely many" primes in the denominator are really just limits of finite products which all go to zero. These are the kinds of issues that arise when you try to interpret an infinite product (or infinite sum for that for that matter) as literally multiplying (or summing) infinitely many terms. $\endgroup$
    – Nico
    Jan 9, 2021 at 3:45
  • 5
    $\begingroup$ Indeed, any valid proof takes a product over finitely many primes, $p_1,\dots,p_N$, then argues that the LHS is the sum over integers only divisible by such primes, and that what's left over tends to zero. $\endgroup$
    – Nico
    Jan 9, 2021 at 3:46
  • $\begingroup$ hi @Nico - if we are to think about the infinite product as a limit, in which case the number of terms in the denominator tends to infinity, then does this mean that the challenge in the linked stack exchange answer is invalid? My apologies for seeking such, perhaps impolite, clarity - I an not a trained mathematician and therefore some concepts I struggle with for longer than most. $\endgroup$
    – Penelope
    Jan 9, 2021 at 15:07
  • $\begingroup$ In the other thread when they talked about fractions with infinite products in the denominators, they were discussing your intuitive understanding only. The whole concept fails to be rigorous by many ways, which was exactly Greg Martin's point. The only way to get anything sensible out of it is throw out the intuitive concepts and go back to the actual definitions. $\endgroup$ Jan 9, 2021 at 16:54

2 Answers 2


The way to prove this rigorously is to show that:

$$\sum_{n\leq N}\frac1{n^s}\leq \prod_{p\leq N}\left(1-1/p^s\right)^{-1}\leq\sum_{n=1}^{\infty}\frac1{n^s}$$ This you can get because the product in the middle is finite, so you can use the argument without worrying about infinitely many primes.

Then use the squeeze theorem as $N\to\infty.$

This technique works only for real $s>1,$ because the inequalities are not true for $s$ complex.

At heart the proof is to let $$A_N=\{n\geq 1\mid n\text{ has no prime factors }>N\}$$

Then you can show that:

$$\prod_{p\leq N}\left(1-1/p^s\right)^{-1} =\sum_{n\in A_N} \frac1{n^s}$$

Then you get the left-hand inequality because $1,2,\dots,N\in A_N$ and the right hand side because $A_N\subseteq \mathbb Z^+.$


By definition, where $p_i$ is the $i^{th}$ prime, $$\prod_{p_i} \left(1-\frac{1}{p_i^s}\right)^{-1}:=\prod_{i=1}^\infty\left(1-\frac{1}{p_i^s}\right)^{-1}:= \lim_N \prod_{i=1}^N\left(1-\frac{1}{p_i^s}\right)^{-1}$$

Now $$\prod_{i=1}^N\left(1-\frac{1}{p_i^s}\right)^{-1} = \prod_{i=1}^N\left(\sum_{e_i=0}^\infty \frac 1{\,p_i^{se_i}\,}\right)$$ And $$\begin{align}\prod_{i=1}^N\left(\sum_{e_i=0}^\infty \frac 1{\,p_i^{se_i}\,}\right) &= \prod_{i=1}^N \lim_{n_i} \sum_{e_i=0}^{n_i} \frac 1{\,p_i^{se_i}\,}\\ &= \lim_{n_1,\dots,n_N} \sum_{e_1=0}^{n_1}\dots\sum_{e_N=0}^{n_N}\frac 1{(p_1^{e_1}p_2^{e_2}\dots p_n^{e_N})^s}\end{align}$$ provided the RHS converges. But since the terms are all positive, the limitend is an increasing function of all its indices. And it is the sum of a finite subsequence of the known convergent sequence $\sum_k \frac 1{k^s}$, so it is bounded above by that value. Hence the RHS side must converge.

Further, it includes a term of the form $\frac 1{k^s}$ for every $k < p_{N+1}$. Thus $$\sum_{k=1}^{p_{N+1}-1} \dfrac1{k^s} \le \lim_{n_1,\dots,n_N} \sum_{e_1=0}^{n_1}\dots\sum_{e_N=0}^{n_N}\frac 1{(p_1^{e_1}p_2^{e_2}\dots p_n^{e_N})^s} \le \sum_{k=1}^\infty \dfrac1{k^s}$$ $$\sum_{k=1}^{p_{N+1}-1} \dfrac1{k^s} \le \prod_{i=1}^N\left(1-\frac{1}{p_i^s}\right)^{-1}\le \sum_{k=1}^\infty \dfrac1{k^s}$$

Since $p_{N+1} \to \infty$ as $N \to \infty$, by the squeeze theorem

$$\sum_{k=1}^\infty \dfrac1{k^s} \le \prod_{i=1}^\infty\left(1-\frac{1}{p_i^s}\right)^{-1} \le \sum_{k=1}^\infty \dfrac1{k^s}$$

That is, $$\prod_{p_i} \left(1-\frac{1}{p_i^s}\right)^{-1} = \sum_{k=1}^\infty \dfrac1{k^s}$$

Note that at one stage in this proof, it required that $$\sum_{k=1}^\infty \dfrac1{k^s}$$ converge. I.e., it only works for $s > 1$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .