### 30th Putnam 1969

Problem B3

The sequence a1, a2, a3, ... satisfies a1a2 = 1, a2a3 = 2, a3a4 = 3, a4a5 = 4, ... . Also, limn→∞an/an+1 = 1. Prove that a1 = √(2/π).

Solution

Let a1 = 1/k. Then we deduce successively that a2 = k, a3 = 2/k, a4 = (3/2) k. By a trivial induction, a2n = (3.5.7 ... 2n-1)/(2.4.6 ... 2n-2) k and a2n+1 = (2.4.6 ... 2n)/(3.5.7 ... 2n-1) 1/k. Hence a2n+1/a2n+2 = (2/1) (2/3) (4/3) (4/5) (6/5) ... (2n/2n-1) 1/k2. We are given that this has limit 1, so (2/1) (2/3) (4/3) (4/5) (6/5) ... (2n/2n-1) has limit k2.

So we need to establish that (2/1) (2/3) (4/3) (4/5) (6/5) (6/7) ... (2n/2n-1) (2n)/(2n+1) ... = p/2. This is the well-known Wallis product. It is usually established by proving the product representation for sin z, but that requires relatively advanced complex analysis, which is outside the Putnam syllabus, so we need a simpler approach. The following is rather unmotivated unless you have seen it before.

Let In = ∫0π/2 sinnx dx. Integrating by parts, we have that In = -∫ sinn-1x d(cos x) = (n-1) In-2 + (n-1) In. Hence In = (n-1)/n In-2. But I0 = π/2, I1 = 1. So we find that I2n = (1/2)(3/4)(5/6) ... (2n-1)/2n π/2, I2n+1 = (2/3)(4/5)(6/7) ... (2n/2n+1).

Now 0 < sin x < 1 on (0, π/2), so I2n-1 < I2n < I2n+1. Dividing by I2n+1, we get (2n+1)/2n > I2n/I2n+1 > 1. So I2n/I2n+1 tends to 1 as n tends to infinity, which estabishes the Wallis product.

© John Scholes
jscholes@kalva.demon.co.uk
14 Jan 2002