Wallis Product

Multiply the rational factors in one at a time and watch the running product swing above and below π2\frac{\pi}{2}, closing in as the swings shrink.

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

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. 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.)

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. 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 error is about 0.06970.0697.
  3. Convergence is slow — five factors still leave a 7% error. (The exact fraction is verified by test.)

6. Series