The distinct points x_{n} are dense in the interval (0, 1). x_{1}, x_{2}, ... , x_{n-1} divide (0, 1) into n sub-intervals, one of which must contain x_{n}. This part is divided by x_{n} into two sub-intervals, lengths a_{n} and b_{n}. Prove that ∑ a_{n}b_{n}(a_{n} + b_{n}) = 1/3.

**Solution**

The trick is to notice that 3a_{n}b_{n}(a_{n} + b_{n}) = (a_{n} + b_{n})^{3} - a_{n}^{3} - b_{n}^{3}. The first n-1 points x_{1}, ... , x_{n-1} divide the interval into n sub-intervals. Let c_{n} be the sum of the cubes of these sub-intervals. Then 3 ∑_{1}^{n-1} a_{i}b_{i}(a_{i} + b_{i}) = 1 - c_{n}. So it is sufficient to prove that c_{n} tends to zero.

Take ε > 0 and < 1. Since the points x_{n} are dense, we can take N so that for n > N all the subintervals are smaller than ε. Then c_{n} < ε^{2} times the sum of the sub-interval lengths = ε^{2} < ε.

© John Scholes

jscholes@kalva.demon.co.uk

5 Feb 2002