### Exercice 1: Finding limits of sequences 1. **Problem:** Find the limit of \( u_n = \frac{\sin(n)}{n^{2} + 3} \). Since \( |\sin(n)| \leq 1 \), we have: $$ |u_n| = \left|\frac{\sin(n)}{n^2 + 3}\right| \leq \frac{1}{n^2 + 3} \rightarrow 0 \quad \text{as } n \to +\infty $$ Thus, \( \lim_{n \to \infty} u_n = 0 \). 2. **Problem:** Find the limit of \( v_n = \frac{1}{E(n\sqrt{2})} \), where \( E \) is the floor function. Since \( E(n\sqrt{2}) \to +\infty \) as \( n \to +\infty \), we have: $$ \lim_{n \to \infty} v_n = \lim_{n \to \infty} \frac{1}{E(n\sqrt{2})} = 0 $$ 3. **Problem:** Find the limit of \( w_n = \frac{1}{n} \). Clearly, $$ \lim_{n \to \infty} w_n = 0 $$ --- ### Exercice 2: Recursive sequence \( (u_n) \) Given: $$ u_0 = 0, \quad u_{n+1} = \sqrt{2 + u_n} $$ 1. **Show:** \( 0 \leq u_n \leq 2 \) for all \( n \in \mathbb{N} \). - Base case \( n=0 \): \( u_0 = 0 \), true. - Assume \( 0 \leq u_n \leq 2 \). Then $$ u_{n+1} = \sqrt{2 + u_n} \geq \sqrt{2 + 0} = \sqrt{2} > 0 $$ Also, $$ u_{n+1} = \sqrt{2 + u_n} \leq \sqrt{2 + 2} = \sqrt{4} = 2 $$ So \( 0 \leq u_{n+1} \leq 2 \). By induction, the inequality holds for all \( n \). 2. **Show:** \( (u_n) \) is increasing and bounded above. - Show \( u_{n+1} \geq u_n \): Since \( u_{n+1} = \sqrt{2 + u_n} \), check if \( u_{n+1} - u_n \geq 0 \). Square both sides where needed to compare: \( u_{n+1} \geq u_n \iff \sqrt{2 + u_n} \geq u_n \) This holds if: $$ 2 + u_n \geq u_n^2 \implies u_n^2 - u_n - 2 \leq 0 $$ The quadratic \( u_n^2 - u_n - 2 = (u_n - 2)(u_n + 1) \) is \( \leq 0 \) for \( u_n \in [-1, 2] \). Since \( 0 \leq u_n \leq 2 \), this is true, so \( u_{n+1} \geq u_n \), i.e., \( (u_n) \) is increasing. It is bounded above by 2 from step 1. 3. **Deduce convergence and find the limit.** Since \( (u_n) \) is increasing and bounded above, it converges. Let \( \ell = \lim_{n \to \infty} u_n \). Taking limit in the recursion: $$ \ell = \sqrt{2 + \ell} \implies \ell^2 = 2 + \ell \implies \ell^2 - \ell - 2 = 0 $$ Solve quadratic: $$ \ell = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2} $$ So \( \ell = 2 \) or \( \ell = -1 \). Since \( u_n \geq 0 \), \( \ell = 2 \). --- ### Exercice 3: Sequences \( (u_n), (v_n), (w_n) \) Given: $$ \begin{cases} u_0 = 0, \ u_{n+1} = \frac{u_n + 2v_n}{3} \end{cases} \quad \begin{cases} v_0 = 11, \ v_{n+1} = \frac{u_n + 4v_n}{5} \end{cases} w_n = v_n - u_n $$ 1. **Study monotonicity of \( (w_n) \).** Calculate \( w_{n+1} \): $$ w_{n+1} = v_{n+1} - u_{n+1} = \frac{u_n + 4v_n}{5} - \frac{u_n + 2v_n}{3} $$ Find common denominator 15: $$ w_{n+1} = \frac{3(u_n + 4v_n) - 5(u_n + 2v_n)}{15} = \frac{3u_n + 12v_n - 5u_n - 10v_n}{15} = \frac{-2u_n + 2v_n}{15} = \frac{2}{15}(v_n - u_n) = \frac{2}{15} w_n $$ Hence: $$ w_{n+1} = \frac{2}{15} w_n $$ Since \( \frac{2}{15} \in (0,1) \), \( (w_n) \) is a geometric sequence with ratio \( \frac{2}{15} \). Thus, \( (w_n) \) is monotonically decreasing if \( w_n > 0 \) or increasing if \( w_n < 0 \), but generally convergent. 2. **Limit of \( (w_n) \).** Since \( w_0 = v_0 - u_0 = 11 - 0 = 11 \), and ratio \( r=\frac{2}{15} \), $$ \lim_{n \to \infty} w_n = w_0 \times 0 = 0 $$ 3. **Limits of even and odd subsequences \( w_{2n} \) and \( w_{2n+1} \).** Since \( w_n = 11 \left( \frac{2}{15} \right)^n \), both subsequences converge to zero as \( n \to \infty \). 4. **Show sequences \( (u_n) \) and \( (v_n) \) are adjacent.** Recall: $$ w_n = v_n - u_n \to 0 $$ Also, \( u_n \leq v_n \) since \( w_n \geq 0 \). Given \( w_n \to 0 \) and \( u_n \) is increasing (proof omitted here but can be shown similarly), and \( v_n = u_n + w_n \) is decreasing since \( w_n \to 0 \), the two sequences are respectively increasing and decreasing and their difference \( w_n \) tends to zero. By definition, \( (u_n) \) and \( (v_n) \) are adjacent sequences. Final answers: - \( \lim u_n = 0 \) in Exercice 1, \( 0 \) for \( u_n \) and \( v_n \), and the defined limits for \( w_n \). - \( (u_n) \) in Exercice 2 converges to 2. - \( w_n \to 0 \) in Exercice 3 and sequences \( u_n, v_n \) are adjacent.