The real numbers a0, a1, a2, ... satisfy 1 = a0 <= a1 ≤ a2 <= ... . b1, b2, b3, ... are defined by bn = sum for k = 1 to n of (1 - ak-1/ak)/√ak.
(a) Prove that 0 ≤ bn < 2.
(b) Given c satisfying 0 ≤ c < 2, prove that we can find an so that bn > c for all sufficiently large n.
Solution
(a) Each term of the sum is non-negative, so bn is non-negative. Let ck = √ak. Then the kth term = (1 - ak-1/ak)/√ak = ck-12/ck (1/ak-1 - 1/ak) = ck-12/ck (1/ck-1 + 1/ck)(1/ck-1 - 1/ck). But ck-12/ck (1/ck-1 + 1/ck) ≤ 2, so the kth term ≤ 2(1/ck-1 - 1/ck). Hence bn <= 2 - 2/cn < 2.
(b) Let ck = dk, where d is a constant > 1, which we will choose later. Then the kth term is (1 - 1/d2)1/dk, so bn = (1 - 1/d2)(1 - 1/dn+1)/(1 - 1/d) = (1 + 1/d)(1 - 1/dn+1). Now take d sufficiently close to 1 that 1 + 1/d > c, and then take n sufficiently large so that (1 + 1/d)(1 - 1/dn+1) > c.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
6 Oct 1998
Last updated/corrected 14 Mar 03