Putnam 1969/B3 solution

30th Putnam 1969

Problem B3

The sequence a₁, a₂, a₃, ... satisfies a₁a₂ = 1, a₂a₃ = 2, a₃a₄ = 3, a₄a₅ = 4, ... . Also, lim_n→∞a_n/a_n+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, a_2n = (3.5.7 ... 2n-1)/(2.4.6 ... 2n-2) k and a_2n+1 = (2.4.6 ... 2n)/(3.5.7 ... 2n-1) 1/k. Hence a_2n+1/a_2n+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 I_n = ∫₀π/2 sinⁿx dx. Integrating by parts, we have that I_n = -∫ sin^n-1x d(cos x) = (n-1) I_n-2 + (n-1) I_n. Hence I_n = (n-1)/n I_n-2. But I₀ = π/2, I₁ = 1. So we find that I_2n = (1/2)(3/4)(5/6) ... (2n-1)/2n π/2, I_2n+1 = (2/3)(4/5)(6/7) ... (2n/2n+1).

Now 0 < sin x < 1 on (0, π/2), so I_2n-1 < I_2n < I_2n+1. Dividing by I_2n+1, we get (2n+1)/2n > I_2n/I_2n+1 > 1. So I_2n/I_2n+1 tends to 1 as n tends to infinity, which estabishes the Wallis product.

30th Putnam 1969