Let $a_0 = 1998$ and $a_{n+1} =\frac{a_n^2}{a_n +1}$ for each nonnegative integer $n$. Prove that $[a_n] = 1994- n$ for $0 \le n \le 1000$
Source: Slovenia TST 1998 p6
Tags: Sequence, recurrence relation, algebra
Let $a_0 = 1998$ and $a_{n+1} =\frac{a_n^2}{a_n +1}$ for each nonnegative integer $n$. Prove that $[a_n] = 1994- n$ for $0 \le n \le 1000$