3rd Irish 1990 Problem 9 Let a n = 2 cos(t/2 n ) - 1. Simplify a 1 a 2 ... a n and deduce that it tends to (2 cos t + 1)/3.
We have cos 2x = 2 cos 2 x - 1, so 2 cos 2x + 1 = 4 cos 2 x - 1 = (2 cos x + 1)(2 cos x - 1). Put b k = 2 cos(t/2 k ) + 1. Then a k = b k-1 /b k . Hence a 1 a 2 ... a n = (2 cos t + 1)/(2 cos(t/2 n ) + 1). As n → ∞, t/2 n → 0, so cos(t/2 n ) → 1 and hence the product tends to (2 cos t + 1)/3. Thanks to Suat Namli 3rd Irish 1990 © John Scholes jscholes@kalva.demon.co.uk 26 February 2004 Last updated/corrected 26 Feb 04