The Euler product, and why it is not a coincidence

5 September 2026

Euler’s product formula is usually presented as a startling identity. It is more useful to read it as a restatement of unique factorisation, written in a language where analysis can get at it.

The statement

For $Re (s) > 1$ ,

\begin{matrix} (1) & ζ (s) = \sum_{n = 1}^{\infty} \frac{1}{n^{s}} = \prod_{p prime} \frac{1}{1 - p^{- s}} . \end{matrix}

Both sides converge absolutely there, which is what makes the rearrangement below legitimate rather than merely suggestive.

Where the product comes from

Expand each factor as a geometric series and multiply out:

\begin{aligned} (2) & \prod_{p \leq N} \frac{1}{1 - p^{- s}} & = \prod_{p \leq N} (1 + p^{- s} + p^{- 2 s} + \dots) \\ (3) & = \sum_{\begin{matrix} n \geq 1 \\ p ∣ n ⟹ p \leq N \end{matrix}} \frac{1}{n^{s}} . \end{aligned}

The second line is unique factorisation in disguise: every $n$ whose prime factors are all at most $N$ appears exactly once, because it has exactly one factorisation.

Letting $N \to \infty$ recovers $(1)$ . The error term is bounded by $\sum_{n > N} n^{- σ}$ with $σ = Re (s) > 1$ , which vanishes.

The consequence worth remembering

Taking $s \to 1^{+}$ makes the left-hand side of $(1)$ diverge. A finite set of primes would give a finite product, so the primes must be infinite — Euler’s analytic proof of Euclid’s theorem.

More precisely, comparing growth rates gives

\begin{matrix} (4) & \sum_{p \leq x} \frac{1}{p} = \log \log x + M + O (\frac{1}{\log x}), \end{matrix}

where $M \approx 0.2615$ is the Meissel–Mertens constant. Estimate $(4)$ says the primes thin out, but only just: slowly enough that their reciprocals still diverge.