Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-09-24 (gpt-6-sol)
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A short Euler-product truncation already numerically approximates zeta at 2

Example

Using the first four primes,

∏p∈{2,3,5,7}11−p−2=1225768≈1.59505,

while

ζ(2)=π26≈1.64493.

Facts & Assumptions

Given: The Euler product and the independent Basel sum.

[L1]

For Re⁡s>1, ζ(s)=∏p(1−p−s)−1 (The Riemann zeta function has its Euler product on the half-plane Re⁡s>1).

[L2]

The Basel sum is ∑n≥1n−2=π2/6 (The Basel sum is pi squared over six by a residue computation), and this convergent Dirichlet series is ζ(2) (The Riemann zeta function on the half-plane Re⁡s>1).

Verification

technique · direct
1.1L1givenalgebra

Evaluating the Euler factors from [L1] at s=2 gives (1−14)−1(1−19)−1(1−125)−1(1−149)−1=43⋅98⋅2524⋅4948=1225768.

2.1step 1.1L2algebra∎

By [L2], the target value is π2/6≈1.64493. Comparing with step 1.1 shows that even this short prime truncation already lands within about five hundredths of ζ(2).

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources