Math  /  Algebra

QuestionExercice ( 5 pts ) On considère la suite UU définie par {U0=23Un+1=3Un+22Un+3;nIN\left\{\begin{array}{c}U_{0}=\frac{2}{3} \\ U_{n+1}=\frac{3 U_{n}+2}{2 U_{n}+3} ; \forall n \in I N\end{array}\right.
1. Calculer U1;U2\boldsymbol{U}_{\mathbf{1}} ; \boldsymbol{U}_{\mathbf{2}}
2. Monter que nIN\forall n \in I N on a : 0Un10 \leq U_{n} \leq 1
3. On pose Vn=Un1NUn+1V_{n}=\frac{U_{n}-1^{N}}{U_{n}+1} a. Monter que la suite (Vn)\left(V_{n}\right) est une suite géométrique b. Calculer Vn\boldsymbol{V}_{\boldsymbol{n}} puis Un\boldsymbol{U}_{\boldsymbol{n}} en fonction de nn c. Calculer Sn=k=0nVnS_{n}=\sum_{k=0}^{n} V_{n} d. Calculer limVn;limUn\lim V_{n} ; \lim U_{n} et limSn\lim S_{n}

Studdy Solution
Calculons les limites.
1. limnVn=limn15n+1=0\lim_{n \to \infty} V_n = \lim_{n \to \infty} -\frac{1}{5^{n+1}} = 0.
2. limnUn=limn115n+11+15n+1=1\lim_{n \to \infty} U_n = \lim_{n \to \infty} \frac{1 - \frac{1}{5^{n+1}}}{1 + \frac{1}{5^{n+1}}} = 1.
3. limnSn=limn14(115n+1)=14\lim_{n \to \infty} S_n = \lim_{n \to \infty} -\frac{1}{4} (1 - \frac{1}{5^{n+1}}) = -\frac{1}{4}.

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord