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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ii/paper-4/1h/solution

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Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 1H / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

Expand the finite Euler product into convergent geometric series. Every positive integer

n \leq x

has all its prime factors at most

x

, so

P (x)^{- 1} = \prod_{p \leq x} \sum_{k \geq 0} p^{- k} = \sum_{n \geq 1 all prime factors of n \leq x} \frac{1}{n} \geq \sum_{n \leq x} \frac{1}{n} .

The harmonic series diverges, hence $P (x) \to 0$ . If the sum of reciprocal primes were finite, then the inequality

- lo g (1 - t) \leq 2 t

for

0 \leq t \leq 1/2

would give

- lo g P (x) = \sum_{p \leq x} - lo g (1 - 1/ p) \leq 2 \sum_{p} \frac{1}{p} < \infty,

uniformly in

x

, contradicting

P (x) \to 0

. Thus the sum of prime reciprocals diverges. Unique factorization justifies the expansion; no estimate for the distribution of primes is required. This is the Euler proof that the sum of reciprocals of primes diverges.

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