bac-s-maths 2025 Q3

bac-s-maths · France · bac-spe-maths__metropole-sept_j1 Sequences and series, recurrence and convergence Convergence proof and limit determination
Exercise 3
The purpose of this exercise is to study the convergence of two sequences towards the same limit.
Part A
Consider the function $f$ defined on $[2;+\infty[$ by $$f(x) = \sqrt{3x-2}$$
  1. Justify the elements of the variation table below:
    $x$2$+\infty$
    $+\infty$
    $f(x)$
    2

    We admit that the sequence $(u_n)$ satisfying $u_0 = 6$ and, for all natural number $n$, $u_{n+1} = f(u_n)$ is well defined.
    1. [a.] Prove by induction that, for all natural number $n$: $2 \leqslant u_{n+1} \leqslant u_n \leqslant 6$.
    2. [b.] Deduce that the sequence $(u_n)$ converges.
  2. We call $\ell$ the limit of $(u_n)$.
    We admit that it is a solution of the equation $f(x) = x$. Determine the value of $\ell$.
  3. Consider the rank function written below in Python language.
    We recall that $\operatorname{sqrt}(x)$ returns the square root of the number $x$.
    \begin{verbatim} from math import * def rang(a) : u=6 n=0 while u >= a : u = sqrt(3*u - 2) n = n+1 return n \end{verbatim}
    1. [a.] Why can we affirm that rang(2.000001) returns a value?
    2. [b.] For which values of the parameter $a$ does the instruction rang($a$) return a result?

Part B
We admit that the sequence $(v_n)$ satisfying $v_0 = 6$ and, for all natural number $n$, $v_{n+1} = 3 - \dfrac{2}{v_n}$ is well defined.
  1. Calculate $v_1$.
  2. For all natural number $n$, we admit that $v_n \neq 2$ and we set: $$w_n = \frac{v_n - 1}{v_n - 2}$$
    1. [a.] Prove that the sequence $(w_n)$ is geometric with ratio 2 and specify its first term $w_0$.
    2. [b.] We admit that, for all natural number $n$, $$w_n - 1 = \frac{1}{v_n - 2}$$ Deduce that, for all natural number $n$, $$v_n = 2 + \frac{1}{1{,}25 \times 2^n - 1}$$
    3. [c.] Calculate the limit of $(v_n)$.
  3. Determine the smallest natural number $n$ for which $v_n < 2{,}01$ by solving the inequality.

Part C
Using the previous parts, determine the smallest integer $N$ such that for all $n \geqslant N$, the terms $v_n$ and $u_n$ belong to the interval $]1{,}99;2{,}01[$.
\textbf{Exercise 3}

The purpose of this exercise is to study the convergence of two sequences towards the same limit.

\textbf{Part A}

Consider the function $f$ defined on $[2;+\infty[$ by
$$f(x) = \sqrt{3x-2}$$

\begin{enumerate}
  \item Justify the elements of the variation table below:

\begin{center}
\begin{tabular}{|c|lll|}
\hline
$x$ & 2 & $+\infty$ & \\
\hline
 & & $+\infty$ & \\
$f(x)$ & & & \\
 & 2 & & \\
\hline
\end{tabular}
\end{center}

We admit that the sequence $(u_n)$ satisfying $u_0 = 6$ and, for all natural number $n$, $u_{n+1} = f(u_n)$ is well defined.

  \item \begin{enumerate}
    \item[a.] Prove by induction that, for all natural number $n$: $2 \leqslant u_{n+1} \leqslant u_n \leqslant 6$.
    \item[b.] Deduce that the sequence $(u_n)$ converges.
  \end{enumerate}
  \item We call $\ell$ the limit of $(u_n)$.

We admit that it is a solution of the equation $f(x) = x$. Determine the value of $\ell$.

  \item Consider the rank function written below in Python language.

We recall that $\operatorname{sqrt}(x)$ returns the square root of the number $x$.

\begin{verbatim}
from math import *
def rang(a) :
    u=6
    n=0
    while u >= a :
        u = sqrt(3*u - 2)
        n = n+1
    return n
\end{verbatim}

    \begin{enumerate}
      \item[a.] Why can we affirm that rang(2.000001) returns a value?
      \item[b.] For which values of the parameter $a$ does the instruction rang($a$) return a result?
    \end{enumerate}
\end{enumerate}

\textbf{Part B}

We admit that the sequence $(v_n)$ satisfying $v_0 = 6$ and, for all natural number $n$, $v_{n+1} = 3 - \dfrac{2}{v_n}$ is well defined.

\begin{enumerate}
  \item Calculate $v_1$.
  \item For all natural number $n$, we admit that $v_n \neq 2$ and we set:
$$w_n = \frac{v_n - 1}{v_n - 2}$$
    \begin{enumerate}
      \item[a.] Prove that the sequence $(w_n)$ is geometric with ratio 2 and specify its first term $w_0$.
      \item[b.] We admit that, for all natural number $n$,
$$w_n - 1 = \frac{1}{v_n - 2}$$
Deduce that, for all natural number $n$,
$$v_n = 2 + \frac{1}{1{,}25 \times 2^n - 1}$$
      \item[c.] Calculate the limit of $(v_n)$.
    \end{enumerate}
  \item Determine the smallest natural number $n$ for which $v_n < 2{,}01$ by solving the inequality.
\end{enumerate}

\textbf{Part C}

Using the previous parts, determine the smallest integer $N$ such that for all $n \geqslant N$, the terms $v_n$ and $u_n$ belong to the interval $]1{,}99;2{,}01[$.
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