TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04

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Landau's theorem for Dirichlet series with nonnegative coefficients

Statement

Let $D (s) = \sum_{n \geq 1} a_{n} n^{- s}$ with $a_{n} \geq 0$ for every $n$ , and assume its abscissa of convergence $σ_{c}$ is finite. Then $s = σ_{c}$ is a singular point of the holomorphic function defined by $D$ on $ℜ s > σ_{c}$ .

Facts & Assumptions

Given: A Dirichlet series $D (s) = \sum a_{n} n^{- s}$ with $a_{n} \geq 0$ and finite abscissa $σ_{c}$ .

[L1]

The abscissa is defined through right-half-plane convergence (The convergence and absolute-convergence abscissae of a Dirichlet series).

[L2]

On every half-plane of absolute convergence, the series and all its derivatives converge locally uniformly (Convergence at one point of a Dirichlet series forces local uniform convergence on the open half-plane to its right, Absolute convergence at one point forces absolute and locally uniform convergence on closed half-planes to the right).

Proof

technique · contradiction

1.1

Suppose $D$ were holomorphic on a disc centered at $σ_{c}$ . Choose a real point $σ_{1} > σ_{c}$ inside that disc and a radius $r > σ_{1} - σ_{c}$ still contained in the disc. By [L2], for every $m \geq 0$ , $\frac{(- 1)^{m}}{m!} D^{(m)} (σ_{1}) = \sum_{n \geq 1} a_{n} \frac{(\log n)^{m}}{m!} n^{- σ_{1}},$ and every coefficient on the right is nonnegative.

L2assume-contra

2.1

The Taylor series of $D$ at $σ_{1}$ therefore has nonnegative coefficients: $D (s) = \sum_{m \geq 0} c_{m} (σ_{1} - s)^{m}, c_{m} \geq 0.$ Because the disc radius exceeds $σ_{1} - σ_{c}$ , this series converges at some real point $σ_{*} < σ_{c}$ . Evaluating there and using the displayed formula for $c_{m}$ gives $\sum_{n \geq 1} a_{n} n^{- σ_{*}} < \infty .$ So the Dirichlet series converges at $σ_{*}$ , contradicting the definition of $σ_{c}$ in [L1].

L1step 1.1

3.1

Hence $σ_{c}$ cannot be a regular point of the holomorphic continuation: it is a singular point.

step 2.1discharge-contradiction∎

Depends on

Used by

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Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Kiran S. Kedlaya, Notes on Analytic Number Theory, Theorem 2.4 (standard reference, not scraped)
Leonard Tomczak, Analytic Number Theory, Theorem 3.6 (standard reference, not scraped)