The purpose of this exercise is to study the sequence $(u_n)$ defined by the value of its first term $u_1$ and, for every natural number $n$ greater than or equal to 1, by the relation:
$$u_{n+1} = (n+1)u_n - 1$$

\textbf{Part A}

\begin{enumerate}
  \item Verify, by detailing the calculation, that if $u_1 = 0$ then $u_4 = -17$.
  \item Copy and complete the algorithm below so that by first entering in $U$ a value of $u_1$ it calculates the terms of the sequence $(u_n)$ from $u_2$ to $u_{13}$.

For $N$ going from 1 to 12
$$U \leftarrow$$
End For

  \item This algorithm was executed for $u_1 = 0.7$ then for $u_1 = 0.8$.

Here are the values obtained.

\begin{center}
\begin{tabular}{|l|l|}
\hline
For $u_1 = 0.7$ & For $u_1 = 0.8$ \\
\hline
0.4 & 0.6 \\
\hline
0.2 & 0.8 \\
\hline
-0.2 & 2.2 \\
\hline
-2 & 10 \\
\hline
-13 & 59 \\
\hline
-92 & 412 \\
\hline
-737 & 3295 \\
\hline
-6634 & 29654 \\
\hline
-66341 & 296539 \\
\hline
-729752 & 3261928 \\
\hline
-8757025 & 39143135 \\
\hline
-113841326 & 508860754 \\
\hline
\end{tabular}
\end{center}

What appears to be the limit of this sequence if $u_1 = 0.7$? And if $u_1 = 0.8$?
\end{enumerate}

\textbf{Part B}

We consider the sequence $(I_n)$ defined for every natural number $n$, greater than or equal to 1, by:
$$I_n = \int_0^1 x^n \mathrm{e}^{1-x} \mathrm{~d}x$$
We recall that the number e is the value of the exponential function at 1, that is to say that $\mathrm{e} = \mathrm{e}^1$.

\begin{enumerate}
  \item Prove that the function $F$ defined on the interval $[0;1]$ by $F(x) = (-1-x)\mathrm{e}^{1-x}$ is an antiderivative on the interval $[0;1]$ of the function $f$ defined on the interval $[0;1]$ by $f(x) = x\mathrm{e}^{1-x}$.
  \item Deduce that $I_1 = \mathrm{e} - 2$.
  \item It is admitted that, for every natural number $n$ greater than or equal to 1, we have:
$$I_{n+1} = (n+1)I_n - 1.$$
Use this formula to calculate $I_2$.
  \item a. Justify that, for every real number $x$ in the interval $[0;1]$ and for every natural number $n$ greater than or equal to 1, we have: $0 \leqslant x^n \mathrm{e}^{1-x} \leqslant x^n \mathrm{e}$.\\
b. Justify that: $\int_0^1 x^n \mathrm{e} \, \mathrm{d}x = \dfrac{\mathrm{e}}{n+1}$.\\
c. Deduce that, for every natural number $n$ greater than or equal to 1, we have: $0 \leqslant I_n \leqslant \dfrac{\mathrm{e}}{n+1}$.\\
d. Determine $\lim_{n \rightarrow +\infty} I_n$.
\end{enumerate}

\textbf{Part C}

In this part, we denote by $n!$ the number defined, for every natural number $n$ greater than or equal to 1, by: $1! = 1$, $2! = 2 \times 1$, and if $n \geqslant 3$: $n! = n \times (n-1) \times \ldots \times 1$.\\
And, more generally: $(n+1)! = (n+1) \times n!$

\begin{enumerate}
  \item Prove by induction that, for every natural number $n$ greater than or equal to 1, we have:
$$u_n = n! \left(u_1 - \mathrm{e} + 2\right) + I_n$$
We recall that, for every natural number $n$ greater than or equal to 1, we have:
$$u_{n+1} = (n+1)u_n - 1 \quad \text{and} \quad I_{n+1} = (n+1)I_n - 1.$$
  \item It is admitted that: $\lim_{n \rightarrow +\infty} n! = +\infty$.\\
a. Determine the limit of the sequence $(u_n)$ when $u_1 = 0.7$.\\
b. Determine the limit of the sequence $(u_n)$ when $u_1 = 0.8$.
\end{enumerate}