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Wallis's product: pi over two is the limit of the finite Wallis products

Statement

For m∈N, define the finite Wallis product

Wm:=∏k=1m(2k)2(2k−1)(2k+1),

with W0=1. Then

lim⁡m→∞Wm=π2.

This limit is the meaning of Wallis's infinite product for π/2.

Facts & Assumptions

Given: The finite products Wm and the Wallis integrals In.

[L1]

The Wallis integrals have the displayed even and odd product forms, and I2m/I2m+1→1 (Wallis integrals satisfy the two-step recurrence, closed forms, and the adjacent-integral squeeze).

[L2]

A finite product in a monoid has empty product equal to the identity and obeys the recursion that adjoins its last factor (The product g0g1⋯gn−1 of a finite list in a monoid, by recursion, with the empty product (n=0) equal to the identity).

[L3]

Products and quotients of convergent real sequences have the corresponding limits when the limiting denominator is nonzero (Algebra of limits: sums, scalar multiples, products and quotients).

Proof

technique · direct
1.1

Substituting the two product formulas of [L1] and collecting matching factors gives I2mI2m+1=π2∏k=1m(2k−1)(2k+1)(2k)2=π/2Wm.

L1L2algebra
2.1

At m=0, step 1.1 reads I0/I1=π/2=(π/2)/1, so the empty-product boundary agrees with the identity. At m=1, it reads I2/I3=3π/8=(π/2)/(4/3), so the first nonempty product agrees as well. Both checks are separate from the limiting assertion.

step 1.1L1L2algebra
2.2

By [L1], the left side of step 1.1 tends to 1. Since every Wm is positive, rearranging gives Wm=(π/2)/(I2m/I2m+1), and [L3] yields Wm→π/2.

step 1.1L1L3
3.1

Thus the finite products, not an undefined completed multiplication, converge to π/2.

step 2.2∎

Depends on

Used by

Dependency tree · two levels

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Sources