 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-25/3/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

- \frac{ζ ^{'} ( s )}{ζ ( s )} = \sum_{p} \sum_{m \geq 1} \frac{l o g p}{p ^{m s}} = \sum_{n \geq 1} \frac{Λ ( n )}{n ^{s}}, ℜ s > 1.

The differentiated sum is absolutely convergent, since

Λ (n) \leq lo g n

. Taking the real parts at the three heights gives

- 3 \frac{ζ ^{'} ( σ )}{ζ ( σ )} - 4ℜ \frac{ζ ^{'} ( σ + i t )}{ζ ( σ + i t )} - ℜ \frac{ζ ^{'} ( σ + 2 i t )}{ζ ( σ + 2 i t )} = \sum_{n \geq 1} \frac{Λ ( n )}{n ^{σ}} (3 + 4 cos (t lo g n) + cos (2 t lo g n)) \geq 0.

Each summand is nonnegative because its bracket is

2 (1 + cos (t lo g n))^{2}

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