Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 11H / b

Table of contents
- Solution b

Solution

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}} .

Equivalently, Dirichlet convolution becomes multiplication of absolutely convergent Dirichlet series.

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