This is one of the most beautiful identities in mathematics:

$$\sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}}$$

The left side sums over all positive integers. The right side takes a product over all primes. They’re equal — and the reason is the Fundamental Theorem of Arithmetic.

Explore: toggle primes on and off

Each bar shows $1/n^2$. Toggle primes to control which integers appear in the sum — only integers whose prime factors are all “on” contribute. This is the Euler product in action: each prime independently decides which terms appear.

Turn off the prime 2 — all even-numbered bars vanish. Turn off 3 as well — anything divisible by 2 or 3 disappears. With only the prime 2 on, the sum is $1 + 1/4 + 1/16 + \cdots = 1/(1 - 1/4) = 4/3$.

Why it works

Start with the product over primes and expand each factor using the geometric series:

$$\frac{1}{1-p^{-s}} = 1 + \frac{1}{p^s} + \frac{1}{p^{2s}} + \frac{1}{p^{3s}} + \cdots$$

So the product becomes:

$$\left(1 + \frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{8^s} + \cdots\right)\left(1 + \frac{1}{3^s} + \frac{1}{9^s} + \frac{1}{27^s} + \cdots\right)\left(1 + \frac{1}{5^s} + \frac{1}{25^s} + \cdots\right)\cdots$$

When you multiply out this infinite product, you pick one term from each factor — one power of 2, one power of 3, one power of 5, and so on. Each choice gives:

$$\frac{1}{(2^{a_2} \cdot 3^{a_3} \cdot 5^{a_5} \cdots)^s}$$

for some exponents $a_2, a_3, a_5, \ldots \geq 0$. But $2^{a_2} \cdot 3^{a_3} \cdot 5^{a_5} \cdots$ is just a positive integer — and by the Fundamental Theorem of Arithmetic, every positive integer appears exactly once. So the product equals the sum over all $n$:

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

Unique prime factorisation is doing the work here. If factorisation weren’t unique, some integers would appear more than once (or not at all) in the expansion, and the identity would fail.

Euler’s proof of infinite primes

Set $s = 1$. The left side becomes the harmonic series, which diverges. The right side is a product over all primes. If there were only finitely many primes, the product would be finite. But the sum diverges — so there must be infinitely many primes. ∎

This is Euler’s analytic proof of the infinitude of primes (1737). It’s fundamentally different from Euclid’s algebraic proof: it uses the size of the sum to constrain the number of primes.

What the Euler product reveals

The identity $\zeta(s) = \prod_p (1-p^{-s})^{-1}$ means:

The analytic properties of $\zeta$ encode the distribution of primes.

• The pole at $s = 1$ (harmonic series diverges) implies infinitely many primes.
• The rate of divergence near $s = 1$ gives the Prime Number Theorem: $\pi(N) \sim N/\ln N$.
• The zeros of $\zeta$ in the complex plane control the error term in the Prime Number Theorem — how close $\pi(N)$ is to $N/\ln N$.
• The Riemann Hypothesis (all non-trivial zeros on $\text{Re}(s) = 1/2$) would give the sharpest possible bound on this error.

The Euler product transforms questions about the multiplicative structure of integers (primes) into questions about the analytic structure of a function (zeta). This is the founding insight of analytic number theory.

Generalisation: L-functions

Replace $1/n^s$ with $a_n/n^s$ for different sequences $a_n$, and the resulting Dirichlet series often have their own Euler products:

$$L(s) = \sum_{n=1}^{\infty} \frac{a_n}{n^s} = \prod_p F_p(p^{-s})^{-1}$$

where $F_p$ depends on the specific L-function. These generalisations encode information about:

• Primes in arithmetic progressions (Dirichlet L-functions)
• Elliptic curves (Hasse-Weil L-functions)
• Modular forms (associated L-functions)

The Taniyama-Shimura conjecture (now a theorem, proved by Wiles et al.) says that the L-function of every rational elliptic curve is also the L-function of a modular form. This connection — between geometry and symmetry, mediated by L-functions — was the key to proving Fermat’s Last Theorem.