Harper Greene
08/23/2023 · Senior High School
3)- Pour tout \( n \in \mathbb{N}^{*} \), On pose : \( S_{n}=\sum_{k=n+1}^{2 n} \frac{g^{-1}\left(u_{k}\right)}{k} \) a) - Montrer que : \( \left(\forall n \in \mathbb{N}^{*}\right), g^{-1}\left(u_{n+1}\right) \cdot \sum_{k=n+1}^{2 n} \frac{1}{k} \leq S_{n} \leq g^{-1}\left(u_{2 n}\right) \cdot \sum_{k=n+1}^{2 n} \frac{1}{k} \) b)- Déduire la limite de la suite \( \left(S_{n}\right)_{n \in \mathbb{N}^{*}} \) \( \quad\left(\right. \) On admet que \( \left.\lim _{n \rightarrow+\infty} \sum_{k=n+1}^{2 n} \frac{1}{k}=\ln 2\right) \)
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a) Montrer que \( g^{-1}(u_{n+1}) \cdot \sum_{k=n+1}^{2 n} \frac{1}{k} \leq S_{n} \leq g^{-1}(u_{2 n}) \cdot \sum_{k=n+1}^{2 n} \frac{1}{k} \) :
- Estimation inférieure : \( g^{-1}(u_{n+1}) \cdot \sum_{k=n+1}^{2 n} \frac{1}{k} \leq S_n \)
- Estimation supérieure : \( S_n \leq g^{-1}(u_{2 n}) \cdot \sum_{k=n+1}^{2 n} \frac{1}{k} \)
b) Déduire la limite de la suite \( (S_n)_{n \in \mathbb{N}^{*}} \) :
- Si \( g^{-1}(u_n) \) converge vers \( L \), alors \( \lim_{n \to +\infty} S_n = L \cdot \ln 2 \).
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