CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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A boundary line need not have uniform convergence behavior

Statement refuted

If a Dirichlet series converges at one point on the line $ℜ s = σ_{c}$ , then it converges at every point on that line.

Facts & Assumptions

Given: The Dirichlet series $D (s) : = \sum_{k \geq 1} \frac{2^{- k s}}{k},$ equivalently the series $\sum_{n \geq 1} a_{n} n^{- s}$ with $a_{2^{k}} = 1 / k$ and $a_{n} = 0$ otherwise.

[L1]

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

[L2]

The $p = 1$ case of the $p$ -series theorem says that the harmonic series diverges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ), while the alternating-series test applied to $1 / k$ says that the alternating harmonic series converges (The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ ).

[L3]

A geometric series with ratio of modulus $< 1$ converges (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

Counterexample

technique · direct

1.1

If $ℜ s = σ > 0$ , then $\sum_{k \geq 1} ∣ \frac{2^{- k s}}{k} ∣ \leq \sum_{k \geq 1} 2^{- k σ},$ and the right-hand side is a convergent geometric series by [L3]. So $D (s)$ converges for every $ℜ s > 0$ . At $s = 0$ it becomes the harmonic series $\sum_{k \geq 1} 1 / k$ , which diverges by [L2]. Therefore [L1] gives $σ_{c} = 0$ .

L1L2L3givenalgebra

2.1

On the same boundary line, at $s_{*} = \frac{π i}{\log 2}$ one has $2^{- s_{*}} = - 1$ , so $D (s_{*}) = \sum_{k \geq 1} \frac{(- 1)^{k}}{k},$ which converges by [L2]. Thus the line $ℜ s = 0$ contains both the divergent point $s = 0$ and the convergent point $s = s_{*}$ .

L2step 1.1algebra

3.1

Therefore convergence at one boundary point does not force convergence at every point of the abscissa line.

step 1.1step 2.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Leonard Tomczak, Analytic Number Theory, Chapter 3 (standard reference, not scraped)