Theorem: Wallis's Product

{{begin-eqn}}
{{eqn | l = \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1}
      | r = \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots
      | c = 
}}
{{eqn | r = \frac \pi 2
      | c = 
}}
{{end-eqn}}

Proof:
From the Reduction Formula for Integral of Power of Sine, we have:
:$\ds (1): \quad \int \sin^n x \rd x = - \frac 1 n \sin^{n - 1} x \cos x + \frac {n - 1} n \int \sin^{n - 2} x \rd x$
Let $I_n$ be defined as:
:$\ds I_n = \int_0^{\pi / 2} \sin^n x \rd x$
As $\cos \dfrac \pi 2 = 0$ from Shape of Cosine Function, we have from $(1)$ that:
:$(2): \quad I_n = \dfrac {n-1} n I_{n - 2}$
To start the ball rolling, we note that:
:$\ds I_0 = \int_0^{\pi / 2} \rd x = \frac \pi 2$
:$\ds I_1 = \int_0^{\pi / 2} \sin x \rd x = \bigintlimits {-\cos x} 0 {\pi / 2} = 1$
We need to separate the cases where the subscripts are even and odd:
{{begin-eqn}}     
{{eqn | l = I_{2 n}
      | r = \frac {2 n - 1} {2 n} I_{2 n - 2}
      | c = 
}}
{{eqn | r = \frac {2 n - 1} {2 n} \cdot \frac {2 n - 3} {2 n - 2} I_{2 n - 4}
      | c = 
}}
{{eqn | r = \cdots
      | c = 
}}
{{eqn | r = \frac {2 n - 1} {2 n} \cdot \frac {2 n - 3} {2 n - 2} \cdot \frac {2 n - 5} {2 n - 4} \cdots \frac 3 4 \cdot \frac 1 2 I_0
      | c = 
}}
{{eqn | n = A
      | r = \frac {2 n - 1} {2 n} \cdot \frac {2 n - 3} {2 n - 2} \cdot \frac {2 n - 5} {2 n - 4} \cdots \frac 3 4 \cdot \frac 1 2 \cdot \frac \pi 2
}}
{{end-eqn}}
{{begin-eqn}}     
{{eqn | l = I_{2 n+1}
      | r = \frac {2 n} {2 n + 1} I_{2 n - 1}
      | c = 
}}
{{eqn | r = \frac {2 n} {2 n + 1} \cdot \frac {2 n - 2} {2 n - 1} I_{2 n - 3}
      | c = 
}}
{{eqn | r = \cdots
      | c = 
}}
{{eqn | r = \frac {2 n} {2 n + 1} \cdot \frac {2 n - 2} {2 n - 1} \cdot \frac {2 n - 4} {2 n - 3} \cdots \frac 4 5 \cdot \frac 2 3 I_1
      | c = 
}}
{{eqn | n = B
      | r = \frac {2 n} {2 n + 1} \cdot \frac {2 n - 2} {2 n - 1} \cdot \frac {2 n - 4} {2 n - 3} \cdots \frac 4 5 \cdot \frac 2 3
}}
{{end-eqn}}
By Shape of Sine Function, we have that on $0 \le x \le \dfrac \pi 2$:
:$0 \le \sin x \le 1$
Therefore:
:$0 \le \sin^{2 n + 2} x \le \sin^{2 n +1} x \le \sin^{2 n} x$
It follows from Relative Sizes of Definite Integrals that:
:$\ds 0 < \int_0^{\pi / 2} \sin^{2 n + 2} x \rd x \le \int_0^{\pi / 2} \sin^{2 n + 1} x \rd x \le \int_0^{\pi / 2} \sin^{2 n} x \rd x$
That is:
:$(3): \quad 0 < I_{2 n + 2} \le I_{2 n + 1} \le I_{2 n}$
By $(2)$ we have:
:$\dfrac {I_{2 n + 2} } {I_{2 n} } = \dfrac {2 n + 1} {2 n + 2}$
Dividing $(3)$ through by $I_{2n}$ then, we have:
:$\dfrac {2 n + 1} {2 n + 2} \le \dfrac {I_{2 n + 1}} {I_{2 n}} \le 1$
By Squeeze Theorem, it follows that:
:$\dfrac {I_{2 n + 1} } {I_{2 n} } \to 1$ as $n \to \infty$
which is equivalent to:
:$\dfrac {I_{2 n} } {I_{2 n + 1} } \to 1$ as $n \to \infty$
Now we take $(B)$ and divide it by $(A)$ to get:
:$\dfrac {I_{2 n + 1} } {I_{2 n} } = \dfrac 2 1 \cdot \dfrac 2 3 \cdot \dfrac 4 3 \cdot \dfrac 4 5 \cdots \dfrac {2 n} {2 n - 1} \cdot \dfrac {2 n} {2 n + 1} \cdot \dfrac 2 \pi$
So:
:$\dfrac \pi 2 = \dfrac 2 1 \cdot \dfrac 2 3 \cdot \dfrac 4 3 \cdot \dfrac 4 5 \cdots \dfrac {2 n} {2 n - 1} \cdot \dfrac {2 n} {2 n + 1} \cdot \paren {\dfrac {I_{2 n} } {I_{2 n + 1} } }$
Taking the limit as $n \to \infty$ gives the result.
{{qed}}