We consider the sequence ( $u _ { n }$ ) such that $u _ { 0 } = 0$ and for all natural integer $n$ :
$$u _ { n + 1 } = \frac { - u _ { n } - 4 } { u _ { n } + 3 } .$$
We admit that $u _ { n }$ is defined for all natural integer $n$.

\begin{enumerate}
  \item Calculate the exact values of $u _ { 1 }$ and $u _ { 2 }$.
  \item We consider the function term below written incompletely in Python language:
\end{enumerate}

\begin{verbatim}
def terme (n) :
    u = ...
    for i in range(n):
        u = ...
    return(u)
\end{verbatim}

We recall that in Python language, «i in range (n) » means that $i$ varies from 0 to $n - 1$.

Rewrite and complete the box above so that, for all natural integer $n$, the instruction terme (n) returns the value of $u _ { n }$.\\
3. Let the function $f$ defined on $] - 3 ; + \infty [$ by:
$$f ( x ) = \frac { - x - 4 } { x + 3 }$$
Thus, for all natural integer $n$, we have $u _ { n + 1 } = f \left( u _ { n } \right)$.\\
Prove that the function $f$ is strictly increasing on $] - 3 ; + \infty [$.\\
4. Prove by induction that for all natural integer $n$ :
$$- 2 < u _ { n + 1 } \leqslant u _ { n } .$$

\begin{enumerate}
  \setcounter{enumi}{4}
  \item Deduce that the sequence ( $u _ { n }$ ) is convergent.
  \item Let the sequence $\left( v _ { n } \right)$ defined for all natural integer $n$ by:
\end{enumerate}
$$v _ { n } = \frac { 1 } { u _ { n } + 2 }$$
a. Give $v _ { 0 }$.\\
b. Prove that the sequence ( $v _ { n }$ ) is arithmetic with common difference 1 .\\
c. Deduce that for all natural integer $n \geqslant 1$ :
$$u _ { n } = \frac { 1 } { n + 0,5 } - 2 .$$
d. Determine the limit of the sequence $\left( u _ { n } \right)$.