1. **Problem Statement:** Solve Exercise 3 Part I and Exercise 4 Part I, and all of Exercise 5. --- ### Exercise 3 Part I Given sequences: $$\begin{cases} u_0 = a, \\ u_{n+1} = \frac{2u_n + v_n}{3} \end{cases} \quad \text{and} \quad \begin{cases} v_0 = b, \\ v_{n+1} = \frac{2v_n + u_n}{3} \end{cases}$$ with $0 < a \leq b$. **1. Determine $u_1$ and $v_1$:** $$u_1 = \frac{2u_0 + v_0}{3} = \frac{2a + b}{3}$$ $$v_1 = \frac{2v_0 + u_0}{3} = \frac{2b + a}{3}$$ --- ### Exercise 4 Part I Given sequence: $$u_n = n \sqrt{n} = n^{3/2}$$ **Definition of divergence to $+\infty$:** 1) For all $A > 0$, there exists $n_0 \in \mathbb{N}$ such that for all $n \geq n_0$, $u_n > A$. 2) For all $A > 0$, there exists $n_0 \in \mathbb{N}$ such that for all $n \geq n_0$, $u_n < -A$. Since $u_n = n^{3/2} > 0$ for $n > 0$, the sequence diverges to $+\infty$ by definition 1. **Given $u_n \in (10^6, +\infty)$, find $A$ and $n_0$:** Set $A = 10^6$. We want $u_n = n^{3/2} > 10^6$. Solve: $$n^{3/2} > 10^6 \implies n > (10^6)^{2/3} = 10^{4}$$ So choose $n_0 = 10^4 + 1$. For all $n \geq n_0$, $u_n > 10^6$. --- ### Exercise 5 Given sequence $(u_n)$ with: $$|u_{n+1} - u_n| \leq k |u_n - u_{n-1}|, \quad n \in \mathbb{N}^*, \quad k \in (0,1)$$ **1. Show:** $$|u_{n+1} - u_n| \leq k^{n-1} |u_2 - u_1|$$ *Proof by induction:* - Base case $n=1$: $$|u_2 - u_1| \leq k^{0} |u_2 - u_1| = |u_2 - u_1|$$ True. - Assume true for $n$, then for $n+1$: $$|u_{n+2} - u_{n+1}| \leq k |u_{n+1} - u_n| \leq k \cdot k^{n-1} |u_2 - u_1| = k^n |u_2 - u_1|$$ Hence true for all $n$. **2. Prove:** $$|u_p - u_q| \leq \frac{k^{q-1}}{1-k} |u_2 - u_1|, \quad q < p$$ *Proof:* By triangle inequality: $$|u_p - u_q| \leq \sum_{j=q}^{p-1} |u_{j+1} - u_j| \leq \sum_{j=q}^{p-1} k^{j-1} |u_2 - u_1| \leq |u_2 - u_1| \sum_{j=q}^{\infty} k^{j-1}$$ Sum of geometric series: $$\sum_{j=q}^{\infty} k^{j-1} = k^{q-1} \sum_{m=0}^{\infty} k^m = \frac{k^{q-1}}{1-k}$$ Therefore: $$|u_p - u_q| \leq \frac{k^{q-1}}{1-k} |u_2 - u_1|$$ **3. Deduce the nature of $(u_n)$:** Since $k \in (0,1)$, $k^{q-1} \to 0$ as $q \to \infty$, so for any $\varepsilon > 0$, choose $N$ such that: $$\frac{k^{N-1}}{1-k} |u_2 - u_1| < \varepsilon$$ Then for all $p,q \geq N$, $$|u_p - u_q| < \varepsilon$$ Hence $(u_n)$ is a Cauchy sequence and therefore convergent. --- **Final answers:** - Exercise 3 Part I: $u_1 = \frac{2a + b}{3}$, $v_1 = \frac{2b + a}{3}$. - Exercise 4 Part I: $u_n = n^{3/2}$ diverges to $+\infty$ with $A = 10^6$, $n_0 = 10^4 + 1$. - Exercise 5: $(u_n)$ satisfies $$|u_{n+1} - u_n| \leq k^{n-1} |u_2 - u_1|,$$ $$|u_p - u_q| \leq \frac{k^{q-1}}{1-k} |u_2 - u_1|,$$ and is a Cauchy sequence, hence convergent.