Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 4 / b

Table of contents
- Solution b

Solution

For

σ > 1

, Orthogonality of Dirichlet characters gives

\sum_{n \geq 1 n \equiv a (mod q)} \frac{Λ ( n )}{n ^{σ}} = \frac{1}{φ ( q )} \sum_{χ mod q} \overline{χ (a)} (- \frac{L ^{'} ( σ , χ )}{L ( σ , χ )}) .

The principal-character term is

\frac{1}{σ - 1} + O_{q} (1) .

Part (a) makes every nonprincipal logarithmic derivative bounded as

σ \to 1^{+}

, so

\sum_{n \geq 1 n \equiv a (mod q)} \frac{Λ ( n )}{n ^{σ}} = \frac{1}{φ ( q ) ( σ - 1 )} + O_{q} (1) ⟶ \infty.

If the nondecreasing Chebyshev function in an arithmetic progression

ψ (x; q, a) = \sum_{n \leq x n \equiv a (mod q)} Λ (n)

were bounded, the Abel summation formula would keep the displayed Dirichlet series bounded near

σ = 1

. Therefore

ψ (x; q, a) ⟶ \infty.

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