The sequence x_{0}, x_{1}, x_{2}, ... is defined by x_{0} = 1994, x_{n+1} = x_{n}^{2}/(x_{n} + 1). Show that [x_{n} ] = 1994 - n for 0 ≤ n ≤ 998.

**Solution**

x_{n} - x_{n+1} = x_{n}/(x_{n} + 1) = 1 - 1/(x_{n} + 1). Also x_{n} > 0 (by a trivial induction, for example), so x_{n} > x_{n+1}.

Also x_{1} = 1993 + 1/1995 > 1993. Hence x_{2} > 1992 + 1/1995 + 1/1994 > 1992, and by a simple induction x_{n} > (1994 - n) + (1/1995 + 1/1994 + ... + 1/(1996-n) ). So certainly [x_{n}] >= 1994 - n. Also for n <= 997, x_{n} > 997.

We have x_{n} = 1994 - n + (1/1995 + 1/(x_{1} + 1) + ... + 1/(x_{n-1} + 1) ). For n ≤ 998, there are at most 998 terms in the parentheses, each term is at most 1/(x_{997} + 1) ≤ 1/998, and all except the last are less than 1/998. Hence for 0 ≤ n ≤ 998, we have x_{n} < 1994 - n + 1. So [x_{n}] = 1994 - n.

*Comment. This is a weak result. [x _{n}] = 1994 - n actually holds for n up to 1261.*

© John Scholes

jscholes@kalva.demon.co.uk

18 Sep 2002