The sequence a_{1}, a_{2}, a_{3}, ... satisfies a_{1}a_{2} = 1, a_{2}a_{3} = 2, a_{3}a_{4} = 3, a_{4}a_{5} = 4, ... . Also, lim_{n→∞}a_{n}/a_{n+1} = 1. Prove that a_{1} = √(2/π).

**Solution**

Let a_{1} = 1/k. Then we deduce successively that a_{2} = k, a_{3} = 2/k, a_{4} = (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/k^{2}. We are given that this has limit 1, so (2/1) (2/3) (4/3) (4/5) (6/5) ... (2n/2n-1) has limit k^{2}.

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} = ∫_{0}π/2 sin^{n}x dx. Integrating by parts, we have that I_{n} = -∫ sin^{n-1}x d(cos x) = (n-1) I_{n-2} + (n-1) I_{n}. Hence I_{n} = (n-1)/n I_{n-2}. But I_{0} = π/2, I_{1} = 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.

© John Scholes

jscholes@kalva.demon.co.uk

14 Jan 2002