The distinct points xn are dense in the interval (0, 1). x1, x2, ... , xn-1 divide (0, 1) into n sub-intervals, one of which must contain xn. This part is divided by xn into two sub-intervals, lengths an and bn. Prove that ∑ anbn(an + bn) = 1/3.
Solution
The trick is to notice that 3anbn(an + bn) = (an + bn)3 - an3 - bn3. The first n-1 points x1, ... , xn-1 divide the interval into n sub-intervals. Let cn be the sum of the cubes of these sub-intervals. Then 3 ∑1n-1 aibi(ai + bi) = 1 - cn. So it is sufficient to prove that cn tends to zero.
Take ε > 0 and < 1. Since the points xn are dense, we can take N so that for n > N all the subintervals are smaller than ε. Then cn < ε2 times the sum of the sub-interval lengths = ε2 < ε.
© John Scholes
jscholes@kalva.demon.co.uk
5 Feb 2002