ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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Wallis partial products are trapped by adjacent sine-power integrals

Example

For $m \geq 1$ , let

$W_{m} : = \prod_{k = 1}^{m} \frac{4 k^{2}}{4 k^{2} - 1} .$

The first three products are

$W_{1} = \frac{4}{3}, W_{2} = \frac{64}{45}, W_{3} = \frac{256}{175} .$

Facts & Assumptions

Given: A positive natural number $m$ and $I_{n} : = \int_{0}^{π / 2} \sin^{n} x d x$ .

[L1]

The Wallis integrals satisfy $I_{2 m + 1} \leq I_{2 m} \leq I_{2 m - 1},$ $I_{2 m} = \frac{π}{2} \prod_{k = 1}^{m} \frac{2 k - 1}{2 k}, I_{2 m + 1} = \prod_{k = 1}^{m} \frac{2 k}{2 k + 1}$ (Wallis integrals satisfy the two-step recurrence, closed forms, and the adjacent-integral squeeze).

[L2]

The products $W_{m}$ converge to $π / 2$ (Wallis's product: pi over two is the limit of the finite Wallis products).

Verification

technique · direct

1.1

Dividing $I_{2 m + 1} \leq I_{2 m}$ in [L1] by the positive odd-over-even product gives $W_{m} \leq π / 2$ .

givenL1algebra

1.2

Using the formula for $I_{2 m - 1}$ obtained from [L1] with $m - 1$ , the other inequality $I_{2 m} \leq I_{2 m - 1}$ gives $\frac{π}{2} \frac{2 m}{2 m + 1} \leq W_{m} .$

givenL1algebra

2.1

Thus $\frac{π}{2} \frac{2 m}{2 m + 1} \leq W_{m} \leq \frac{π}{2},$ which is the exact finite trap coming from the adjacent integrals.

step 1.1step 1.2

3.1

Multiplication gives the three displayed values; for $m = 1, 2, 3$ , step 2.1 respectively places them in $\frac{π}{3} \leq \frac{4}{3} \leq \frac{π}{2}, \frac{2 π}{5} \leq \frac{64}{45} \leq \frac{π}{2}, \frac{3 π}{7} \leq \frac{256}{175} \leq \frac{π}{2} .$

step 2.1algebra

4.1

The exact bounds in step 2.1 certify every finite product without decimal approximations, while [L2] supplies their limit $π / 2$ .

step 2.1L2∎

Depends on

Used by

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Sources

Imperial College London, History of Mathematics, Problems VI solutions (standard reference, not scraped)