 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-3/1g/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

ϑ (x) = \sum_{p \leq x p prime} lo g p .

Taking natural logarithms in part a gives

2 n lo g 2 - lo g (2 n + 1) \leq 2 n lo g (2 n) + ϑ (2 n),

and hence

ϑ (2 n) \geq 2 n lo g 2 - 2 n lo g (2 n) - lo g (2 n + 1) .

Because

n lo g n / n \to 0

, the right-hand side is at least

n /2

for every sufficiently large integer

n

Now let

x

be sufficiently large and set

n = ⌊ x /2 ⌋

using the floor function. Then

2 n \leq x

and

n \geq x /3

, so the monotonicity of

ϑ

gives

p \leq x p prime \sum lo g p = ϑ (x) \geq ϑ (2 n) \geq \frac{n}{2} \geq \frac{x}{6} \geq \frac{x}{12} .

 New to topics? Read the docs here!