Wallis Product

Rational factors get multiplied in one at a time here. The running product swings around one value, tightening as it goes.

The running product alternates above and below its target. The swings shrink as more factors pile in. So the product settles onto a single value. The surprise is that this value is π divided by 2.

For example, a 🚗 car eases into the center of its lane with small steering corrections to the left and right. Each correction is smaller than the last, and the car steadies in the middle.

π2=∏n=1∞(2n)2(2n−1)(2n+1)\frac{\pi}{2}=\prod_{n=1}^{\infty}\frac{(2n)^2}{(2n-1)(2n+1)}

Examples & Exercises

For n≥2n \ge 2, prove In=∫0π/2sin⁡nx dxI_n = \int_0^{\pi/2}\sin^n x\,dx satisfies In=n−1nIn−2I_n = \frac{n-1}{n}I_{n-2} by integration by parts.

In=n−1nIn−2,I0=π2, I1=1I_n = \frac{n-1}{n}I_{n-2}, \qquad I_0=\frac{\pi}{2},\ I_1=1
This recurrence is the engine of the Wallis product.
Show proof

  1. Assume n≥2n \ge 2 — this is what makes sin⁡n−2x\sin^{n-2}x and In−2I_{n-2} defined and the boundary-term argument below valid. Write In=∫0π/2sin⁡n−1x⋅sin⁡x dxI_n = \int_0^{\pi/2}\sin^{n-1}x\cdot\sin x\,dx and integrate by parts with u=sin⁡n−1xu = \sin^{n-1}x, dv=sin⁡x dxdv = \sin x\,dx: In=[−cos⁡xsin⁡n−1x]0π/2+(n−1)∫0π/2cos⁡2xsin⁡n−2x dxI_n = \left[-\cos x\sin^{n-1}x\right]_0^{\pi/2} + (n-1)\int_0^{\pi/2}\cos^2x\sin^{n-2}x\,dx.

  2. The boundary term vanishes for n≥2n \ge 2. Since cos⁡2x=1−sin⁡2x\cos^2x = 1-\sin^2x, In=(n−1)(In−2−In)I_n = (n-1)(I_{n-2}-I_n).

  3. Rearranging, nIn=(n−1)In−2nI_n = (n-1)I_{n-2}, i.e. In=n−1nIn−2I_n = \frac{n-1}{n}I_{n-2}. (With I0=π2I_0 = \frac{\pi}{2} and I1=1I_1 = 1.)

For n≥1n \ge 1, from sin⁡2n+1x≤sin⁡2nx≤sin⁡2n−1x\sin^{2n+1}x \le \sin^{2n}x \le \sin^{2n-1}x on [0,π2][0,\frac{\pi}{2}], show I2nI2n+1→1\frac{I_{2n}}{I_{2n+1}} \to 1 and derive the Wallis product.

I2nI2n+1=π2∏k=1n(2k−1)(2k+1)(2k)2=π/2Wn→1\frac{I_{2n}}{I_{2n+1}} = \frac{\pi}{2}\prod_{k=1}^{n}\frac{(2k-1)(2k+1)}{(2k)^2} = \frac{\pi/2}{W_n} \to 1
One squeeze sends the product to π/2\pi/2.
Show proof

  1. Take n≥1n \ge 1 (n=0n = 0 would require I−1I_{-1}).

    Integrating gives I2n+1≤I2n≤I2n−1I_{2n+1} \le I_{2n} \le I_{2n-1}, so 1≤I2nI2n+1≤I2n−1I2n+1=2n+12n→11 \le \frac{I_{2n}}{I_{2n+1}} \le \frac{I_{2n-1}}{I_{2n+1}} = \frac{2n+1}{2n} \to 1.

    By the squeeze theorem, I2nI2n+1→1\frac{I_{2n}}{I_{2n+1}} \to 1.

  2. Iterating the recurrence, I2n=π2∏k=1n2k−12kI_{2n} = \frac{\pi}{2}\prod_{k=1}^{n}\frac{2k-1}{2k} and I2n+1=∏k=1n2k2k+1I_{2n+1} = \prod_{k=1}^{n}\frac{2k}{2k+1}. Dividing, I2nI2n+1=π2∏k=1n(2k−1)(2k+1)(2k)2=π/2Wn\frac{I_{2n}}{I_{2n+1}} = \frac{\pi}{2}\prod_{k=1}^{n}\frac{(2k-1)(2k+1)}{(2k)^2} = \frac{\pi/2}{W_n}.

  3. Since the left side tends to 11, Wn→π2W_n \to \frac{\pi}{2}, i.e. ∏n=1∞(2n)2(2n−1)(2n+1)=π2\prod_{n=1}^{\infty}\frac{(2n)^2}{(2n-1)(2n+1)} = \frac{\pi}{2}.

Compute P5=∏n=15(2n)2(2n−1)(2n+1)P_5 = \prod_{n=1}^{5}\frac{(2n)^2}{(2n-1)(2n+1)} exactly as a fraction and compare with π2\frac{\pi}{2}.

P5=6553643659≈1.50109P_5 = \frac{65536}{43659} \approx 1.50109
n=5n = 5: still short of π/2≈1.57080\pi/2 \approx 1.57080.
Show proof

  1. P5=221⋅3⋅423⋅5⋅625⋅7⋅827⋅9⋅1029⋅11=6553643659≈1.50109P_5 = \frac{2^2}{1\cdot3}\cdot\frac{4^2}{3\cdot5}\cdot\frac{6^2}{5\cdot7}\cdot\frac{8^2}{7\cdot9}\cdot\frac{10^2}{9\cdot11} = \frac{65536}{43659} \approx 1.50109.

  2. Since π2≈1.57080\frac{\pi}{2} \approx 1.57080, the absolute error is about 0.06970.0697, and the relative error is 0.06971.57080≈4.4%\frac{0.0697}{1.57080} \approx 4.4\%.

  3. Convergence is slow — five factors still leave about 4.4%4.4\% relative error (absolute error 0.06970.0697). (The exact fraction is verified by test.)

7. Series