The sequence x0, x1, x2, ... is defined by x0 = 1994, xn+1 = xn2/(xn + 1). Show that [xn ] = 1994 - n for 0 ≤ n ≤ 998.
Solution
xn - xn+1 = xn/(xn + 1) = 1 - 1/(xn + 1). Also xn > 0 (by a trivial induction, for example), so xn > xn+1.
Also x1 = 1993 + 1/1995 > 1993. Hence x2 > 1992 + 1/1995 + 1/1994 > 1992, and by a simple induction xn > (1994 - n) + (1/1995 + 1/1994 + ... + 1/(1996-n) ). So certainly [xn] >= 1994 - n. Also for n <= 997, xn > 997.
We have xn = 1994 - n + (1/1995 + 1/(x1 + 1) + ... + 1/(xn-1 + 1) ). For n ≤ 998, there are at most 998 terms in the parentheses, each term is at most 1/(x997 + 1) ≤ 1/998, and all except the last are less than 1/998. Hence for 0 ≤ n ≤ 998, we have xn < 1994 - n + 1. So [xn] = 1994 - n.
Comment. This is a weak result. [xn] = 1994 - n actually holds for n up to 1261.
© John Scholes
jscholes@kalva.demon.co.uk
18 Sep 2002