CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-09-04

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The defining Dirichlet series for zeta diverges at $s = 1$ because it becomes the harmonic series

Statement refuted

The defining series of zeta still converges at $s = 1$ .

Facts & Assumptions

Given: The defining Dirichlet series.

[L1]

On $Re s > 1$ , zeta is defined by $\sum_{n \geq 1} n^{- s}$ (The Riemann zeta function on the half-plane $Re s > 1$ ).

[L2]

The series $\sum_{n \geq 1} 1 / n$ diverges because $p = 1$ is the threshold case (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ).

Counterexample

technique · direct

1.1

At $s = 1$ , the defining series in [L1] becomes $\sum_{n \geq 1} 1 / n$ .

L1given

2.1

By [L2], this is the harmonic series and it diverges. Therefore the defining Dirichlet series does not converge at $s = 1$ .

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

16 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 (standard reference, not scraped)

The defining Dirichlet series for zeta diverges at s=1s=1 because it becomes the harmonic series