Wallis's Product/Original Proof
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Theorem
\(\ds \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1}\) | \(=\) | \(\ds \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac \pi 2\) |
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:
\(\ds I_{2 n}\) | \(=\) | \(\ds \frac {2 n - 1} {2 n} I_{2 n - 2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 n - 1} {2 n} \cdot \frac {2 n - 3} {2 n - 2} I_{2 n - 4}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cdots\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 n - 1} {2 n} \cdot \frac {2 n - 3} {2 n - 2} \cdot \frac {2 n - 5} {2 n - 4} \cdots \frac 3 4 \cdot \frac 1 2 I_0\) | ||||||||||||
\(\text {(A)}: \quad\) | \(\ds \) | \(=\) | \(\ds \frac {2 n - 1} {2 n} \cdot \frac {2 n - 3} {2 n - 2} \cdot \frac {2 n - 5} {2 n - 4} \cdots \frac 3 4 \cdot \frac 1 2 \cdot \frac \pi 2\) |
\(\ds I_{2 n+1}\) | \(=\) | \(\ds \frac {2 n} {2 n + 1} I_{2 n - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 n} {2 n + 1} \cdot \frac {2 n - 2} {2 n - 1} I_{2 n - 3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \cdots\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 n} {2 n + 1} \cdot \frac {2 n - 2} {2 n - 1} \cdot \frac {2 n - 4} {2 n - 3} \cdots \frac 4 5 \cdot \frac 2 3 I_1\) | ||||||||||||
\(\text {(B)}: \quad\) | \(\ds \) | \(=\) | \(\ds \frac {2 n} {2 n + 1} \cdot \frac {2 n - 2} {2 n - 1} \cdot \frac {2 n - 4} {2 n - 3} \cdots \frac 4 5 \cdot \frac 2 3\) |
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.
$\blacksquare$
Source of Name
This entry was named for John Wallis.
Sources
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.12$: Wallis's Product