Here is a sum built from every positive integer:

It has no obvious opinion about primes at all — it happily adds up 4^{-s} and 6^{-s} and 1000^{-s} right alongside the prime terms. And yet it turns out to be exactly equal to an infinite product that mentions only the primes:

This is the Euler product idea, now read as a statement about the zeta function. It is arguably the single most important identity connecting sums and primes in all of mathematics — the precise hinge that lets analysis (limits, calculus, complex functions) reach in and touch arithmetic (which numbers are prime).

The identity

The left side sees every integer; the right side sees only primes. Their equality means \zeta is a faithful encoding of the primes — anything we learn about the function becomes a fact about the primes, and vice versa.

Where does it come from?

The engine underneath is unique factorisation: every integer n \ge 1 is a product of prime powers in exactly one way. Start from the geometric series for a single prime p:

Now multiply one such series together for every prime p, and expand the giant product term by term. Each term picks one power of 2, one power of 3, one power of 5, and so on, and multiplies them — producing exactly 1/n^{s} for some integer n. Because factorisation into primes is unique, every integer n is produced by exactly one combination of choices, so every term 1/n^{s} appears in the expansion exactly once. Collecting them all back up gives precisely \sum_n 1/n^{s} = \zeta(s). Unique factorisation is the whole proof, dressed in analytic clothing.

Making it concrete: where does 1/12^{s} come from?

The derivation above can feel abstract, so let's catch the machinery in the act for one specific number. Since 12 = 2^{2} \cdot 3, the term 1/12^{s} should appear exactly once when the giant product is expanded. Here is where it comes from:

• from the prime-2 factor (1 + 2^{-s} + 2^{-2s} + \cdots), pick the term 2^{-2s} (that's the "use two factors of 2" choice);
• from the prime-3 factor (1 + 3^{-s} + 3^{-2s} + \cdots), pick the term 3^{-s} (the "use one factor of 3" choice);
• from every other prime's factor (5, 7, 11, \ldots), pick the plain 1 (the "use none of this prime" choice).

Multiply the chosen terms together: 2^{-2s} \cdot 3^{-s} = (2^{2}\cdot 3)^{-s} = 12^{-s}. Because 12 factors into primes in exactly one way, no other combination of choices could ever produce it — so this term is picked out neither twice nor never, but exactly once. Repeat this for every integer, and the expanded product is forced to equal the full sum \sum_n 1/n^{s}, term for term.

Even n=1 fits the pattern: pick the plain "1" term from every single prime's factor, and the empty combination of choices multiplies out to exactly 1 = 1^{-s}. Nothing is left over and nothing is missing — which is exactly what "exactly once" demands.

Worked example: watching both sides meet

Euler himself already knew that \zeta(2) = \pi^2/6 \approx 1.6449 (the famous "Basel problem"). Let's check the identity numerically at s=2 by truncating both sides and watching them close in on the same number.

Summing the first 10 terms of the left side:

Multiplying the first few factors of the right side — one for each prime up to 13:

Both are heading toward the same target, \pi^2/6 \approx 1.6449 — and notice the product, built from just six primes, is already closer than the sum of ten terms. A handful of primes carries a surprising amount of the whole sum's weight. Run the code below to see it for yourself, and try raising the cutoffs.

The chart below plots both running totals as the cutoff grows — the sum climbing up through more and more integers, the product climbing through more and more primes. Watch them both squeeze in on the same dashed line from below:

The primes are encoded in the zeros

Taking a logarithm turns the product into a sum over primes, and differentiating relates \zeta to a sum that counts primes (and prime powers). The upshot — Riemann's great insight — is an exact formula for the prime-counting function \pi(x) written in terms of the zeros of \zeta:

Each zero contributes an oscillation; together they correct the smooth estimate into the exact prime count. The primes' apparent randomness is the music of the zeta zeros.

Concretely, the logarithmic derivative -\zeta'(s)/\zeta(s) unpacks — again via the Euler product — into a sum weighted by the von Mangoldt function \Lambda(n), which is \log p when n is a power of the prime p, and 0 otherwise. That single function, summed up, counts primes (with a careful weighting for prime powers) — and it is precisely this sum that Riemann's formula above pins down exactly, zero by zero.

Why this matters

This is the strategy behind every proof of the Prime Number Theorem: control where \zeta can vanish, and you control how the primes are spread. Showing simply that \zeta(s) \ne 0 on the line \Re(s) = 1 is already enough to prove it. Pinning the zeros down exactly is the Riemann Hypothesis.

This is not just a hypothetical strategy — it is exactly how the Prime Number Theorem was finally proved. In 1896, two mathematicians working independently, Jacques Hadamard and Charles Jean de la Vallée Poussin, both showed that \zeta(s) has no zeros on the line \Re(s)=1, and both used it — via the Euler product — to nail down exactly how the primes thin out among the integers. A century of effort chasing this single sum-product identity finally paid off.