 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-4/11h/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Take both functions to be the constant arithmetic function

1

. Then

(1 ⋆ 1) (n) = \sum_{d ∣ n} 1 = τ (n),

so the divisor function is multiplicative by part (a). For

s

with

Re s > 1

, the relevant Dirichlet series are absolutely convergent, so their Cauchy product may be rearranged:

ζ (s)^{2} = a, b \geq 1 \sum \frac{1}{( ab ) ^{s}} = n \geq 1 \sum \frac{1}{n ^{s}} a ∣ n \sum 1 = n \geq 1 \sum \frac{τ ( n )}{n ^{s}} .

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