Let's go through the problem step by step to solve it.

Exercise 4 Overview:

We are given a sequence $(u_n)$ defined by:

• $u_0 = 0$
• $u_{n+1} = \frac{2}{5} u_n + 1$, for $n \in \mathbb{N}$

The tasks are as follows:

1. Show by induction that $u_n \leq \frac{5}{3}$ for all $n \in \mathbb{N}$.

Induction Proof Outline:

• Base case: Show that $u_0 \leq \frac{5}{3}$.
• Inductive hypothesis: Assume $u_n \leq \frac{5}{3}$ holds for some $n = k$.
• Inductive step: Show that $u_{n+1} = \frac{2}{5} u_n + 1 \leq \frac{5}{3}$ under the assumption.

2. a) Show that $u_{n+1} - u_n = -\frac{3}{5} \left( u_n - \frac{5}{3} \right)$.

We need to manipulate the recurrence relation $u_{n+1} = \frac{2}{5} u_n + 1$ to express the difference $u_{n+1} - u_n$ in terms of $u_n$.

b) Deduce that the sequence $(u_n)$ is increasing.

Using the result from part (a), we analyze the sign of $u_{n+1} - u_n$ to determine whether the sequence is increasing.

3. Define $v_n = u_n - \frac{5}{3}$ for all $n \in \mathbb{N}$.

This creates a new sequence $(v_n)$, and the following tasks are based on this definition:

a) Calculate $v_0$.

b) Show that the sequence $(v_n)$ is geometric with a ratio $q = \frac{2}{5}$.

c) Express $v_n$ as a function of $n$, and deduce the general formula for $u_n$.

Let's start solving each step systematically.

Would you like a detailed solution for each part?