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

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The eta series can represent the continued zeta function where the defining Dirichlet series diverges

Statement refuted

Whenever a series represents the zeta continuation, it must be the defining Dirichlet series $\sum_{n \geq 1} n^{- s}$ .

Facts & Assumptions

Given: The point $s = 1 / 2$ .

[L1]

On $Re s > 0$ , $\sum_{n \geq 1} (- 1)^{n - 1} n^{- s} = (1 - 2^{1 - s}) ζ (s)$ (The Dirichlet eta series is holomorphic on $Re s > 0$ and equals the prefactor times zeta there).

[L2]

The series $\sum_{n \geq 1} n^{- 1 / 2}$ diverges because $1 / 2 \leq 1$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ).

Counterexample

technique · direct

1.1

Apply [L1] at $s = 1 / 2$ . Then $\sum_{n \geq 1} \frac{(- 1)^{n - 1}}{\sqrt{n}} = (1 - \sqrt{2}) ζ (1 / 2),$ so the eta series represents the continued zeta value at $1 / 2$ .

L1given

2.1

By [L2], the defining Dirichlet series $\sum_{n \geq 1} n^{- 1 / 2}$ diverges. Therefore step 1.1 gives a point where the continuation is represented by the eta series even though the defining Dirichlet series diverges.

step 1.1L2algebra∎

Depends on

Used by

FALSE: zeta is given by the same Dirichlet series for every complex s other than 1 False statement

Dependency tree · two levels

19 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

Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 6 §2.1 (standard reference, not scraped)