Turn unique factorization into a convergent Euler product

Try

At exponent two, the Euler factor for prime two is 1+1/4+1/16+…=4/3. It collects the reciprocal squares of powers of two. Multiply by the prime-three factor 9/8 and obtain 3/2. The expanded product contains reciprocal squares of exactly those integers whose prime factors are only two and three. Thus the product records arithmetic structure, not just a numerical coincidence between two formulas.

Predict

Predict what happens when more prime factors are included. Every positive integer eventually appears, and it appears exactly once by unique factorization. But an infinite algebraic expansion needs a convergence argument. We use a real exponent s>1 and define ζ(s)=Σ from n=1 to infinity of 1/n^s. Comparison with the integral of x^(−s) shows this sum is finite. At s=1 the harmonic series diverges, so the same finite-limit conclusion cannot be asserted.

Experiment

Inspect the integers represented by a finite-prime product.

Allowed primes	Example supported integers	Euler product at s=2
None	1	1
2	1, 2, 4, 8,…	4/3
2, 3	1, 2, 3, 4, 6, 8, 9,…	3/2

Find an integer absent from the last row’s supported set: five is absent, even though many larger integers are present. Increasing the prime cutoff is not the same operation as truncating the integer sum at a fixed n.

Understand

For a finite set P of primes, expand product over p in P of(1−p^(−s))^(−1). Each factor is a convergent geometric series. The product sums n^(−s) over integers whose prime factors belong to P. All terms are nonnegative. As P grows through primes up to a cutoff, these sums increase and include every fixed integer eventually. Their limit is therefore ζ(s), proving the Euler product. One can avoid abstract convergence terminology by choosing an integer cutoff N: once all primes up to N are included, every term through N appears, while the omitted contribution is bounded by the convergent tail of the full series.

Apply

Dirichlet multiplication follows the same accounting. If Σ∣f(n)∣/n^s and Σ∣g(n)∣/n^s converge, their product can be regrouped by n=ab. The coefficient is the divisor convolution (f∗g)(n)=Σ over d dividing n of f(d)g(n/d). For f=g=1 this gives Στ(n)/n^s=ζ(s)². For f=μ and g=1, absolute convergence follows from ∣μ(n)∣≤1, and the Möbius cancellation sum makes the product one. Hence Σμ(n)/n^s=1/ζ(s). These statements are analytic identities under stated convergence conditions, not merely formal rearrangements.

Continue

As an additional example, Σφ(n)/n^s=ζ(s−1)/ζ(s) for s>2 follows from Σd|nφ(d)=n and absolute convergence usingφ(n)≤n. The stricter exponent range matters. Ordinary generating functions organize coefficients by addition of indices; Dirichlet series organize them by multiplication of indices and divisors. Analytic continuation of ζ, zeros, and prime-distribution theorems are further topics; none follows merely from the elementary product proof. In numerical work report both the cutoff and a justified tail bound rather than treating a short product as an exact zeta value.

Return to the number theory unit to choose a targeted retry. Continue with prove every positive integer is a sum of four squares.

Compare the ordinary generating-function lesson and recurrence methods: their index addition differs from the divisor multiplication used here.