Wallis's Product/Original Proof

Theorem


\prod_{n=1}^{\infty} \frac{2n}{2n-1} \cdot \frac{2n}{2n+1} = \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 = \frac{\pi}{2} $$

Proof
From the Reduction Formula for Integral Power of Sine Function, we have:
 * $$(1) \qquad \int \sin^n x \mathrm{d}{x} = - \frac 1 n \sin ^{n-1} \cos x + \frac {n-1} n \int \sin^{n-2} x \mathrm{d}{x}$$

Let $$I_n$$ be defined as:
 * $$I_n = \int_0^{\pi / 2} \sin^n x \mathrm{d}{x}$$

As $$\cos \frac \pi 2 = 0$$ from Nature of Cosine Function, we have from $$(1)$$ that:
 * $$(2) \qquad I_n = \frac {n-1} n I_{n-2}$$

To start the ball rolling, we note that:
 * $$I_0 = \int_0^{\pi / 2} \mathrm{d}{x} = \frac \pi 2 \qquad \qquad I_1 = \int_0^{\pi / 2} \sin x \mathrm{d}{x} = \left[{- \cos x}\right]_0^{\pi / 2} = 1$$

We need to separate the cases where the subscripts are even and odd:

$$ $$ $$ $$ $$

$$ $$ $$ $$ $$

By Nature of Sine Function, we have that on $$0 \le x \le \frac \pi 2$$ we have that $$0 \le \sin x \le 1$$.

Therefore:
 * $$0 \le \sin^{2n+2} x \le \sin^{2n+1} x \le \sin^{2n} x$$

It follows from ... that:
 * $$0 < \int_0^{\pi/2} \sin^{2n+2} x \mathrm{d} x \le \int_0^{\pi/2} \sin^{2n+1} x \mathrm{d} x \le \int_0^{\pi/2} \sin^{2n} x \mathrm{d} x$$

That is:
 * $$(3) \qquad 0 < I_{2n+2} \le I_{2n+1} \le I_{2n}$$

By $$(2)$$ we have:
 * $$\frac {I_{2n_2}}{I_{2n}} = \frac {2n+1}{2n+2}$$

Dividing $$(3)$$ through by $$I_{2n}$$ then, we have:
 * $$\frac {2n+1}{2n+2} \le \frac {I_{2n+1}}{I_{2n}} \le 1$$

It follows then that:
 * $$\frac {I_{2n+1}}{I_{2n}} \to 1$$ as $$n \to \infty$$

which is equivalent to:
 * $$\frac {I_{2n}}{I_{2n+1}} \to 1$$ as $$n \to \infty$$

Now we take $$(B)$$ and divide it by $$A$$ to get:


 * $$\frac {I_{2n+1}}{I_{2n}} = \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdots \frac {2n} {2n-1} \cdot \frac {2n} {2n+1} \cdot \frac 2 \pi$$

So:
 * $$\frac \pi 2 = \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdots \frac {2n} {2n-1} \cdot \frac {2n} {2n+1} \cdot \left({\frac {I_{2n}}{I_{2n+1}}}\right)$$

Taking the limit as $$n \to \infty$$ gives the result.