SCHWEITZER 2019 Problem 6

Let $d$ be a positive integer and $1 < a \le (d+2)/(d+1)$ . For given $x_0, x_1,\dots, x_d \in (0, a-1)$ , let $x_{k+1} = x_k (a - x_{k-d})$ , $k \ge d$ . Prove that $\lim_{k \to \infty} x_k = a-1$ .