Wallis's Product/Original Proof

Theorem

$ \displaystyle \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:

$\displaystyle (1) \qquad \int \sin^n x \mathrm d x = - \frac 1 n \sin ^{n-1} \cos x + \frac {n-1} n \int \sin^{n-2} x \mathrm d x$

Let $I_n$ be defined as:

$\displaystyle I_n = \int_0^{\pi / 2} \sin^n x \mathrm d x$

As $\displaystyle \cos \frac \pi 2 = 0$ from Shape 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:

$\displaystyle I_0 = \int_0^{\pi / 2} \mathrm d x = \frac \pi 2 \qquad \qquad I_1 = \int_0^{\pi / 2} \sin x \mathrm d x = \left[{- \cos x}\right]_0^{\pi / 2} = 1$

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle I_{2n}\)	\(=\)	\(\displaystyle \frac {2n-1}{2n} I_{2n-2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {2n-1}{2n} \cdot \frac {2n-3}{2n-2} I_{2n-4}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \cdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {2n-1}{2n} \cdot \frac {2n-3}{2n-2} \cdot \frac {2n-5}{2n-4} \cdots \frac 3 4 \cdot \frac 1 2 I_0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {2n-1}{2n} \cdot \frac {2n-3}{2n-2} \cdot \frac {2n-5}{2n-4} \cdots \frac 3 4 \cdot \frac 1 2 \cdot \frac \pi 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	$(A)$

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle I_{2n+1}\)	\(=\)	\(\displaystyle \frac {2n}{2n+1} I_{2n-1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {2n}{2n+1} \cdot \frac {2n-2}{2n-1} I_{2n-3}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \cdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {2n}{2n+1} \cdot \frac {2n-2}{2n-1} \cdot \frac {2n-4}{2n-3} \cdots \frac 4 5 \cdot \frac 2 3 I_1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {2n}{2n+1} \cdot \frac {2n-2}{2n-1} \cdot \frac {2n-4}{2n-3} \cdots \frac 4 5 \cdot \frac 2 3\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	$(B)$

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

Therefore:

$\displaystyle 0 \le \sin^{2n+2} x \le \sin^{2n+1} x \le \sin^{2n} x$

It follows from Relative Sizes of Definite Integrals that:

$\displaystyle 0 < \int_0^{\pi/2} \sin^{2n+2} x \ \mathrm d x \le \int_0^{\pi/2} \sin^{2n+1} x \ \mathrm d x \le \int_0^{\pi/2} \sin^{2n} x \ \mathrm d x$

That is:

$\displaystyle (3) \qquad 0 < I_{2n+2} \le I_{2n+1} \le I_{2n}$

By $(2)$ we have:

$\displaystyle \frac {I_{2n_2}}{I_{2n}} = \frac {2n+1}{2n+2}$

Dividing $(3)$ through by $I_{2n}$ then, we have:

$\displaystyle \frac {2n+1}{2n+2} \le \frac {I_{2n+1}}{I_{2n}} \le 1$

It follows then that:

$\displaystyle \frac {I_{2n+1}}{I_{2n}} \to 1$ as $n \to \infty$

which is equivalent to:

$\displaystyle \frac {I_{2n}}{I_{2n+1}} \to 1$ as $n \to \infty$

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

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

So:

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

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

$\blacksquare$

Sources

• George F. Simmons: Calculus Gems (1992), Chapter $\text {B}.12$