1. Soit \( \left(u_{n}\right) \) la suite définie par: \( \left\{\begin{array}{l}u_{0}=-3, \\ u_{n+1}=\frac{2 u_{n}+5}{u_{n}+6}, \quad n \in \mathbb{N} \text {. }\end{array}\right. \) (a) Montrer que pour tout \( n \in \mathbb{N}, \frac{1}{u_{n+1}-1}=\frac{7}{u_{n}-1}+1 \) (b) En déduire la limite de \( \left(u_{n}\right) \). (c) Déterminer le plus petit entier \( n_{0} \), tel que pour tout \( n \geq n_{0}, u_{n}>0,99 \).

Step 1 ：\( u_{n+1} = \frac{2u_n + 5}{u_n + 6} \)

Step 2 ：\( \frac{1}{u_{n+1} - 1} = \frac{2u_n + 5}{u_n + 6} - 1 \)

Step 3 ：\( \frac{1}{u_{n+1} - 1} = \frac{2u_n + 5 - u_n - 6}{u_n + 6} \)

Step 4 ：\( \frac{1}{u_{n+1} - 1} = \frac{u_n - 1}{u_n + 6} \)

Step 5 ：\( \frac{1}{u_{n+1} - 1} + 1 = \frac{u_n - 1}{u_n + 6} + \frac{u_n + 6}{u_n + 6} \)

Step 6 ：\( \frac{1}{u_{n+1} - 1} + 1 = \frac{7(u_n - 1) + u_n + 6}{u_n + 6} \)

Step 7 ：\( \frac{1}{u_{n+1} - 1} + 1 = \frac{7(u_n - 1) + u_n + 6}{u_n + 6} \) \Rightarrow (a)

Step 8 ：\(u_n \rightarrow 1 \) as \( n \rightarrow \infty \) \Rightarrow (b)

Step 9 ：Iterating from \(u_0\): \(u_0 =-3\), \(u_1 = 1/3\), \(u_2 =3/7\), \(u_3 = 47/53\), \(u_4 = 329/335 \)

Step 10 ：Since \(u_4 > 0.99\), smallest \(n_0 = 4\) \Rightarrow (c)