---
tag: 0C3PN4075R
kind: statement
role: theorem
name: Euler product formula
msc: [11M06]
uses: [0SN0MAEKYX, 084HH1NDCB]
sources:
  - url: https://en.wikipedia.org/wiki/Euler_product
    locator: Examples
---

## Statement

Let $s$ be a complex number with real part $\operatorname{Re}(s) > 1$, and let $\zeta(s) = \sum_{n=1}^{\infty} n^{-s}$ be the Riemann zeta function.
Then
$$\zeta(s) = \prod_{p} \frac{1}{1 - p^{-s}},$$
where the product runs over all [[prime numbers|0SN0MAEKYX]] $p$ and converges absolutely.

## Remarks

The proof expands each factor as a geometric series $1 + p^{-s} + p^{-2s} + \cdots$. The [[fundamental theorem of arithmetic|084HH1NDCB]] shows that the product of these series contains each term $n^{-s}$ exactly one time.
The same method gives a product formula $\sum_n f(n) n^{-s} = \prod_p \sum_{k \ge 0} f(p^k) p^{-ks}$ for each [[multiplicative function|0AP80TT660]] $f$ when the series converges absolutely.
Since $\zeta(s) \to \infty$ as $s \to 1^+$, the formula gives a proof of [[Euclid's theorem|0292EPPPT0]].

## Explanations

### L1 human

Multiply $(1 + \tfrac12 + \tfrac14 + \cdots)(1 + \tfrac13 + \tfrac19 + \cdots)(1 + \tfrac15 + \cdots) \cdots$ and expand.
Each fraction $1/n$ appears one time, because $n$ is a product of prime powers in exactly one way.
The formula with the exponent $s$ is the same idea with $1 / n^s$.