Exercise 4 — 6 points\\
Theme: sequences, functions\\
Let $(u_n)$ be the sequence defined by $u_0 = -1$ and, for every natural number $n$:
$$u_{n+1} = 0.9u_n - 0.3.$$
\begin{enumerate}
  \item a. Prove by induction that, for all $n \in \mathbb{N}$, $u_n = 2 \times 0.9^n - 3$.\\
b. Deduce that for all $n \in \mathbb{N}$, $-3 < u_n \leq -1$.\\
c. Prove that the sequence $(u_n)$ is strictly decreasing.\\
d. Prove that the sequence $(u_n)$ converges and specify its limit.
  \item We propose to study the function $g$ defined on $]-3; -1]$ by:
$$g(x) = \ln(0.5x + 1.5) - x.$$
a. Justify all the information given by the variations table of function $g$ (limits, variations, image of $-1$).\\
b. Deduce that the equation $g(x) = 0$ has exactly one solution which we will denote $\alpha$ and for which we will give an interval of amplitude $10^{-3}$.
  \item In the rest of the exercise, we consider the sequence $(v_n)$ defined for all $n \in \mathbb{N}$ by:
$$v_n = \ln(0.5u_n + 1.5).$$
a. Using the formula given in question 1.a., prove that the sequence $v$ is arithmetic with common difference $\ln(0.9)$.\\
b. Let $n$ be a natural number.\\
Prove that $u_n = v_n$ if and only if $g(u_n) = 0$.\\
c. Prove that there is no rank $k \in \mathbb{N}$ for which $u_k = \alpha$.\\
d. Deduce that there is no rank $k \in \mathbb{N}$ for which $v_k = u_k$.
\end{enumerate}