 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-4/1i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

For

x \geq 2

, expand the finite Euler product as a product of geometric series:

P (x) := \prod_{p \leq x} (1 - \frac{1}{p})^{- 1} = \sum_{n \geq 1 p ∣ n \Rightarrow p \leq x} \frac{1}{n} .

1 \leq n \leq x

has all its prime factors at most

x

, so

P (x) \geq \sum_{n \leq x} \frac{1}{n} .

The harmonic series diverges, and therefore

p \prod (1 - \frac{1}{p})^{- 1} = \infty.

Suppose instead that

\sum_{p} 1/ p

converged. For

0 < t \leq 1/2

0 \leq - lo g (1 - t) - t = \sum_{j \geq 2} \frac{t ^{j}}{j} \leq \frac{t ^{2}}{1 - t} \leq 2 t^{2} .

Since

\sum_{p} 1/ p^{2}

converges, this would make

lo g P (x) = \sum_{p \leq x} - lo g (1 - \frac{1}{p})

converge to a finite limit, contradicting the divergence of

P (x)

p \sum \frac{1}{p} = \infty.

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