We consider the sequence $(u_n)$ defined by $u_0 = 8$ and, for every natural number $n$,
$$u_{n+1} = \frac{6u_n + 2}{u_n + 5}.$$

\begin{enumerate}
  \item Calculate $u_1$.
  \item Let $f$ be the function defined on the interval $[0; +\infty[$ by:
$$f(x) = \frac{6x + 2}{x + 5}.$$
Thus, for every natural number $n$, we have: $u_{n+1} = f(u_n)$.

  a. Prove that the function $f$ is strictly increasing on the interval $[0; +\infty[$. Deduce that for every real number $x > 2$, we have $f(x) > 2$.

  b. Prove by induction that, for every natural number $n$, we have $u_n > 2$.

  \item We admit that, for every natural number $n$, we have:
$$u_{n+1} - u_n = \frac{(2 - u_n)(u_n + 1)}{u_n + 5}.$$

  a. Prove that the sequence $(u_n)$ is decreasing.

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

  \item We define the sequence $(v_n)$ for every natural number by:
$$v_n = \frac{u_n - 2}{u_n + 1}.$$

  a. Calculate $v_0$.

  b. Prove that $(v_n)$ is a geometric sequence with common ratio $\dfrac{4}{7}$.

  c. Determine, by justifying, the limit of $(v_n)$. Deduce the limit of $(u_n)$.

  \item We consider the Python function threshold below, where A is a real number strictly greater than 2.
\begin{verbatim}
def seuil(A):
    n = 0
    u = 8
    while u > A:
        u = (6*u + 2)/(u + 5)
        n = n + 1
    return n
\end{verbatim}
Give, without justification, the value returned by the command \texttt{seuil(2.001)} then interpret this value in the context of the exercise.
\end{enumerate}