The purpose of this exercise is the study of the sequence $(u_n)$ defined by its first term $u_1 = \frac{3}{2}$ and the recurrence relation: $u_{n+1} = \frac{n u_n + 1}{2(n+1)}$.
Part A - Algorithms and conjectures
To calculate and display the term $u_9$ of the sequence, a student proposes the algorithm below. He forgot to complete two lines.
Variables
$n$ is a natural integer, $u$ is a real number
Initialization
Assign to $n$ the value 1, Assign to $u$ the value 1.5
Treatment
While $n < 9$, Assign to $u$ the value $\cdots$, Assign to $n$ the value $\cdots$, End While
Output
Display the variable $u$
Copy and complete the two lines of the algorithm where there are ellipses.
How would this algorithm need to be modified so that it calculates and displays all terms of the sequence from $u_2$ to $u_9$?
With this modified algorithm, the following results were obtained, rounded to the nearest ten-thousandth:
n
1
2
3
4
5
6
$\ldots$
99
100
$u_n$
1.5
0.625
0.375
0.2656
0.2063
0.1693
$\ldots$
0.0102
0.0101
In light of these results, conjecture the direction of variation and convergence of the sequence $(u_n)$.
Part B - Mathematical study
We define an auxiliary sequence $(v_n)$ by: for all integer $n \geqslant 1, v_n = n u_n - 1$.
Show that the sequence $(v_n)$ is geometric; specify its common ratio and its first term.
Deduce that, for all natural integer $n \geqslant 1$, we have: $u_n = \frac{1 + (0.5)^n}{n}$.
Determine the limit of the sequence $(u_n)$.
Justify that, for all integer $n \geqslant 1$, we have: $u_{n+1} - u_n = -\frac{1 + (1 + 0.5n)(0.5)^n}{n(n+1)}$. Deduce the direction of variation of the sequence $(u_n)$.
Part C - Return to algorithms
Inspired by part A, write an algorithm to determine and display the smallest integer $n$ such that $u_n < 0.001$.
The purpose of this exercise is the study of the sequence $(u_n)$ defined by its first term $u_1 = \frac{3}{2}$ and the recurrence relation: $u_{n+1} = \frac{n u_n + 1}{2(n+1)}$.
\section*{Part A - Algorithms and conjectures}
To calculate and display the term $u_9$ of the sequence, a student proposes the algorithm below. He forgot to complete two lines.
\begin{center}
\begin{tabular}{|l|l|}
\hline
Variables & $n$ is a natural integer, $u$ is a real number \\
\hline
Initialization & Assign to $n$ the value 1, Assign to $u$ the value 1.5 \\
\hline
Treatment & While $n < 9$, Assign to $u$ the value $\cdots$, Assign to $n$ the value $\cdots$, End While \\
\hline
Output & Display the variable $u$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}
\item Copy and complete the two lines of the algorithm where there are ellipses.
\item How would this algorithm need to be modified so that it calculates and displays all terms of the sequence from $u_2$ to $u_9$?
\item With this modified algorithm, the following results were obtained, rounded to the nearest ten-thousandth:
\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|c|c|c|}
\hline
n & 1 & 2 & 3 & 4 & 5 & 6 & $\ldots$ & 99 & 100 \\
\hline
$u_n$ & 1.5 & 0.625 & 0.375 & 0.2656 & 0.2063 & 0.1693 & $\ldots$ & 0.0102 & 0.0101 \\
\hline
\end{tabular}
\end{center}
In light of these results, conjecture the direction of variation and convergence of the sequence $(u_n)$.
\end{enumerate}
\section*{Part B - Mathematical study}
We define an auxiliary sequence $(v_n)$ by: for all integer $n \geqslant 1, v_n = n u_n - 1$.
\begin{enumerate}
\item Show that the sequence $(v_n)$ is geometric; specify its common ratio and its first term.
\item Deduce that, for all natural integer $n \geqslant 1$, we have: $u_n = \frac{1 + (0.5)^n}{n}$.
\item Determine the limit of the sequence $(u_n)$.
\item Justify that, for all integer $n \geqslant 1$, we have: $u_{n+1} - u_n = -\frac{1 + (1 + 0.5n)(0.5)^n}{n(n+1)}$. Deduce the direction of variation of the sequence $(u_n)$.
\end{enumerate}
\section*{Part C - Return to algorithms}
Inspired by part A, write an algorithm to determine and display the smallest integer $n$ such that $u_n < 0.001$.