x_{1} < x_{2} < x_{3} < ... is a sequence of positive reals such that lim x_{n}/n = 0. Is it true that we can find arbitrarily large N such that all of (x_{1} + x_{2N-1}), (x_{2} + x_{2N-2}), (x_{3} + x_{2N-3}), ... , (x_{N-1} + x_{N+1}) are less than 2 x_{N}?

**Solution**

Answer: yes.

Given 0 < k < x_{1}, take N to be the largest n which maximises (x_{n} - nk). k is chosen so that (x_{1} - k) is positive. Since x_{n}/n tends to zero, only finitely many (x_{n} - nk) are positive, so N is well defined. Now x_{N-m} - (N - m)k ≤ x_{N} - Nk and x_{N+m} - (N + m)k < x_{N} - Nk. Adding gives x_{N-m} + x_{N+m} < 2x_{N}. So we are home provided we can show that this process generates arbitrarily large N.

The trick is to take k small. In particular, take k < (x_{N+1} - x_{N})/(N+1). Then x_{N+1} - (N+1)k > x_{N}. A fortiori it is greater than all of x_{i} - ki for i ≤ N. So the value corresponding to this k is bigger than N, which completes the proof.

© John Scholes

jscholes@kalva.demon.co.uk

16 Dec 2001