Bonjour
Membre1999
Exercice 89
a) q = 1/2
Croissance de la suite (Sn)
[tex]S_{n+1}-S_n=(u_0+u_1+...+u_n+u_{n+1})-(u_0+u_1+...+u_n)\\\\S_{n+1}-S_n=u_0+u_1+...+u_n+u_{n+1}-u_0-u_1-...-u_n\\\\S_{n+1}-S_n=(u_0-u_0)+(u_1-u_1)+...+(u_n-u_n)+u_{n+1}\\\\S_{n+1}-S_n=0+0+...+0+u_{n+1}\\\\S_{n+1}-S_n=u_{n+1}\\\\S_{n+1}-S_n=u_0\times q^{n+1}\\\\S_{n+1}-S_n=1\times(\dfrac{1}{2})^{n+1}\\\\S_{n+1}-S_n=(\dfrac{1}{2})^{n+1}\ \textgreater \ 0\\\\\Longrightarrow\boxed{S_{n+1}\ \textgreater \ S_n}[/tex]
Par conséquent, la suite (Sn) est strictement croissante.
Limite de la suite (Sn)
[tex]S_n=u_0\times\dfrac{1-q^n}{1-q}\\\\\\S_n=1\times\dfrac{1-(\dfrac{1}{2})^n}{1-\dfrac{1}{2}}\\\\\\S_n=\dfrac{1-(\dfrac{1}{2})^n}{\dfrac{1}{2}}\\\\\\S_n=[1-(\dfrac{1}{2})^n]\times\dfrac{2}{1}\\\\\\\Longrightarrow\boxed{S_n=2[1-(\dfrac{1}{2})^n]}[/tex]
Or [tex]\lim\limits_{n\to+\infty}(\dfrac{1}{2})^n=0\ \ car\ \ 0\ \textless \ \dfrac{1}{2}\ \textless \ 1[/tex]
D'où
[tex]\lim\limits_{n\to+\infty}S_n=2[1-0]\\\\\\\boxed{\lim\limits_{n\to+\infty}S_n=2}[/tex]
b) q = 1/3
Croissance de la suite (Sn)
[tex]S_{n+1}-S_n=(u_0+u_1+...+u_n+u_{n+1})-(u_0+u_1+...+u_n)\\\\S_{n+1}-S_n=u_0+u_1+...+u_n+u_{n+1}-u_0-u_1-...-u_n\\\\S_{n+1}-S_n=(u_0-u_0)+(u_1-u_1)+...+(u_n-u_n)+u_{n+1}\\\\S_{n+1}-S_n=0+0+...+0+u_{n+1}\\\\S_{n+1}-S_n=u_{n+1}\\\\S_{n+1}-S_n=u_0\times q^{n+1}\\\\S_{n+1}-S_n=1\times(\dfrac{1}{3})^{n+1}\\\\S_{n+1}-S_n=(\dfrac{1}{3})^{n+1}\ \textgreater \ 0\\\\\Longrightarrow\boxed{S_{n+1}\ \textgreater \ S_n}[/tex]
Par conséquent, la suite (Sn) est strictement croissante.
Limite de la suite (Sn)
[tex]S_n=u_0\times\dfrac{1-q^n}{1-q}\\\\\\S_n=1\times\dfrac{1-(\dfrac{1}{3})^n}{1-\dfrac{1}{3}}\\\\\\S_n=\dfrac{1-(\dfrac{1}{3})^n}{\dfrac{2}{3}}\\\\\\S_n=[1-(\dfrac{1}{3})^n]\times\dfrac{3}{2}\\\\\\\Longrightarrow\boxed{S_n=\dfrac{3}{2}\times[1-(\dfrac{1}{3})^n]}[/tex]
Or [tex]\lim\limits_{n\to+\infty}(\dfrac{1}{3})^n=0\ \ car\ \ 0\ \textless \ \dfrac{1}{3}\ \textless \ 1[/tex]
D'où
[tex]\lim\limits_{n\to+\infty}S_n=\dfrac{3}{2}\times[1-0]\\\\\\\boxed{\lim\limits_{n\to+\infty}S_n=\dfrac{3}{2}}[/tex]
