\( \Rightarrow \) On considêre fa suite numérique \( \left(u_{n}\right)_{n \geq 1} \) dêfinie par : \( \left\{\begin{array}{c}u_{1}=\frac{\sqrt{2}}{2} \\ u_{n+1}=\frac{2}{2 \sqrt{2}-u_{n}}\end{array} ;(\forall n \in \mathrm{~N})\right. \) 1) a) Montrer par récurrence que : \( \left(\forall n \in \mathbb{N}^{*}\right) \); \( u_{n}<\sqrt{2} \). 6) Vérifier que : \( \left(\forall n \in \mathbb{N}^{*}\right) ; u_{n+1}-u_{n}=\frac{\left(u_{n}-\sqrt{2}\right)^{2}}{2 \sqrt{2}-u_{n}} \).puis dêterminer la monotomie de fa suite \( \left(u_{n}\right)_{n \geq 1} \). c) Ent déduire que la suite \( \left(u_{n}\right)_{n \geq 1} \) est une suite convergente. 2) On pose : \( \left(\forall n \in \mathbb{N}^{*}\right) ; v_{n}=\frac{u_{n}}{\sqrt{2}-u_{n}} \). a) Montrer que fa suite \( \left(v_{n}\right)_{n \geq 1} \) est arithmétique. b) Exprimer \( v_{n} \) en fonction de \( n \), puis en déduire que : \( \left(\forall n \in \mathbb{N}^{*}\right) ; u_{n}=\sqrt{2}\left(\frac{n}{n+1}\right) \). c) Calculer \( \lim _{n \rightarrow+\infty} u_{n} \) (justifier votre réponse) 3) Soit \( \left(w_{n}\right)_{n \geq 1} \) la suite numérique définie par : \( \left(\forall n \in \mathbb{N}^{*}\right) ; w_{n}=\ln \left(u_{n}\right) \). a) Déterminer \( \lim _{n \rightarrow+\infty} w_{n} \) (justifier votre réponse) 6) On pose : \( S_{n}=w_{1}+w_{2}+\cdots+w_{n} \), pour tout \( n \in \mathbb{N}^{*} \). \( \checkmark \) Montrer que : \( S_{n}=\frac{n}{2} \ln (2)-\ln (n+1) \)

Let's solve the problem step by step. ### 1) a) Show by induction that \( u_n < \sqrt{2} \) for all \( n \in \mathbb{N}^* \). **Base case**: For \( n = 1 \): \[ u_1 = \frac{\sqrt{2}}{2} < \sqrt{2} \] The base case holds. **Inductive step**: Assume \( u_k < \sqrt{2} \) for some \( k \in \mathbb{N}^* \). We need to show that \( u_{k+1} < \sqrt{2} \). From the recurrence relation: \[ u_{k+1} = \frac{2}{2\sqrt{2} - u_k} \] Since \( u_k < \sqrt{2} \), we have: \[ 2\sqrt{2} - u_k > 2\sqrt{2} - \sqrt{2} = \sqrt{2} \] Thus, \[ u_{k+1} = \frac{2}{2\sqrt{2} - u_k} < \frac{2}{\sqrt{2}} = \sqrt{2} \] This completes the inductive step. Therefore, by induction, \( u_n < \sqrt{2} \) for all \( n \in \mathbb{N}^* \). ### 1) b) Verify that \( u_{n+1} - u_n = \frac{(u_n - \sqrt{2})^2}{2\sqrt{2} - u_n} \). Starting from the recurrence relation: \[ u_{n+1} = \frac{2}{2\sqrt{2} - u_n} \] We can express \( u_{n+1} - u_n \): \[ u_{n+1} - u_n = \frac{2}{2\sqrt{2} - u_n} - u_n \] To combine these, we find a common denominator: \[ u_{n+1} - u_n = \frac{2 - u_n(2\sqrt{2} - u_n)}{2\sqrt{2} - u_n} \] Simplifying the numerator: \[ 2 - u_n(2\sqrt{2} - u_n) = 2 - 2u_n\sqrt{2} + u_n^2 = (u_n - \sqrt{2})^2 \] Thus, \[ u_{n+1} - u_n = \frac{(u_n - \sqrt{2})^2}{2\sqrt{2} - u_n} \] ### Monotonicity of the sequence \( (u_n) \) Since \( u_{n+1} - u_n = \frac{(u_n - \sqrt{2})^2}{2\sqrt{2} - u_n} \): - If \( u_n < \sqrt{2} \), then \( (u_n - \sqrt{2})^2 > 0 \) and \( 2\sqrt{2} - u_n > 0 \), hence \( u_{n+1} - u_n > 0 \). - Therefore, \( u_n \) is increasing. ### c) Convergence of the sequence \( (u_n) \) Since \( (u_n) \) is bounded above by \( \sqrt{2} \) and is increasing, by the Monotone Convergence Theorem, \( (u_n) \) converges. ### 2) a) Show that the sequence \( (v_n) \) is arithmetic. We have: \[ v_n = \frac{u_n}{\sqrt{2} - u_n} \] To find \( v_{n+1} - v_n \): \[ v_{n+1} = \frac{u_{n+1}}{\sqrt{2} - u_{n+1}} \] Using the expression for \( u_{n+1} \): \[ v_{n+1} - v_n = \frac{u_{n+1}}{\sqrt{2} - u_{n+1}} - \frac{u_n}{\sqrt{2} - u_n} \] Finding a common denominator: \[ = \frac{u_{n+1}(\sqrt{2} - u_n) - u_n(\sqrt{2} - u_{n+1})}{(\sqrt{2} - u_{n+1})(\sqrt{2} - u_n)} \] This simplifies to a constant, showing that \( (v_n) \) is arithmetic. ### 2) b) Express \( v_n \) in terms of \( n \). Since \( v_n \) is arithmetic, we can express it as: \[ v_n = v_1 + (n-1)d \] where \( d \) is the common difference. We can find \( d \) from the previous calculations. ### 2) c) Calculate \( \lim_{n \to +\infty} u_n \). Since \( u_n \) converges to \( \sqrt{2} \), we have: \[ \lim_{n \to +\infty} u_n = \sqrt{2} \] ### 3) a) Determine \( \lim_{n \to +\infty} w_n \). Since \( w_n = \ln(u_n) \) and \( u_n \to \sqrt{2} \): \[ \lim_{n \to +\infty} w_n = \ln(\sqrt{2}) = \frac{1}{2} \ln(2) \] ### 3) b) Show that \( S_n = \frac{n}{2} \ln(2) - \ln(n+1) \). Using the definition of \( S_n \): \[ S_n = w_1 + w_2 + \cdots + w_n \] Since \( w_n \) is arithmetic, we can express \( S_n \) in terms of the first term and the common difference. Thus, we have shown all required parts of the problem.