1. **Problem statement:** Consider the sequence $(U_n)$ defined by $U_0=1$ and $U_{n+1} = \frac{9}{6 - U_n}$ for all $n \in \mathbb{N}$. Show that $0 < U_n < 3$ for all $n$. 2. **Formula and rules:** We will use mathematical induction and properties of inequalities to prove the bounds on $U_n$. 3. **Base case:** For $n=0$, $U_0=1$ which satisfies $0 < 1 < 3$. 4. **Inductive step:** Assume $0 < U_n < 3$. Then since $U_n < 3$, $6 - U_n > 3 > 0$, so the denominator is positive. Also, $U_n > 0$ implies $6 - U_n < 6$. 5. Compute $U_{n+1} = \frac{9}{6 - U_n}$. Since $6 - U_n > 3$, $U_{n+1} < \frac{9}{3} = 3$. Also, since $6 - U_n < 6$, $U_{n+1} > \frac{9}{6} = 1.5 > 0$. 6. Therefore, $0 < U_{n+1} < 3$. By induction, $0 < U_n < 3$ for all $n$.