Exercise 3The purpose of this exercise is to study the convergence of two sequences towards the same limit.
Part AConsider the function $f$ defined on $[2;+\infty[$ by $$f(x) = \sqrt{3x-2}$$
- 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.
- [a.] Prove by induction that, for all natural number $n$: $2 \leqslant u_{n+1} \leqslant u_n \leqslant 6$.
- [b.] Deduce that the sequence $(u_n)$ converges.
- 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 $\ell$.
- 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}
- [a.] Why can we affirm that rang(2.000001) returns a value?
- [b.] For which values of the parameter $a$ does the instruction rang($a$) return a result?
Part BWe 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.
- Calculate $v_1$.
- For all natural number $n$, we admit that $v_n \neq 2$ and we set: $$w_n = \frac{v_n - 1}{v_n - 2}$$
- [a.] Prove that the sequence $(w_n)$ is geometric with ratio 2 and specify its first term $w_0$.
- [b.] We admit that, for all natural number $n$, $$w_n - 1 = \frac{1}{v_n - 2}$$ Deduce that, for all natural number $n$, $$v_n = 2 + \frac{1}{1{,}25 \times 2^n - 1}$$
- [c.] Calculate the limit of $(v_n)$.
- Determine the smallest natural number $n$ for which $v_n < 2{,}01$ by solving the inequality.
Part CUsing 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 $]1{,}99;2{,}01[$.