Exercice 1
Calculer les limites : \(u_n = \frac{3n^2 - n}{2n^2 + 1}\), \(v_n = \frac{2^n - 3^n}{2^n + 3^n}\), \(w_n = \frac{\sin n}{n}\).
Correction
\(u_n\to\frac32\). \(v_n = \frac{(\frac23)^n - 1}{(\frac23)^n + 1}\to-1\). \(|w_n|\le\frac1n\to0\).
Exercice 2
Soit \(u_0 = 0\) et \(u_{n+1} = \frac{2u_n + 3}{u_n + 4}\). Montrer que \(0\le u_n < 1\), que \((u_n)\) est croissante, et calculer sa limite.
Correction
Si \(0\le u_n < 1\) : \(u_{n+1} - 1 = \frac{u_n - 1}{u_n + 4} < 0\) et \(u_{n+1} > 0\). \(u_{n+1} - u_n = \frac{(1 - u_n)(u_n + 3)}{u_n + 4} > 0\). Limite : \(\ell = \frac{2\ell + 3}{\ell + 4}\Leftrightarrow\ell^2 + 2\ell - 3 = 0\), \(\ell = 1\).
Exercice 3
Soient \(u_n = \sum_{k=1}^{n}\frac{1}{k^2}\) et \(v_n = u_n + \frac1n\). Montrer que \((u_n)\) et \((v_n)\) sont adjacentes.
Correction
\(u_{n+1} - u_n = \frac{1}{(n+1)^2} > 0\). \(v_{n+1} - v_n = \frac{1}{(n+1)^2} + \frac{1}{n + 1} - \frac1n = \frac{n - (n + 1)}{n(n+1)^2} < 0\). \(v_n - u_n = \frac1n\to0\).
Exercice 4
Soit \(u_n = \sum_{k=1}^{n}\frac{1}{n + k}\). Montrer que \(\frac12\le u_n < 1\) et que \((u_n)\) est croissante.
Correction
Chaque terme est entre \(\frac{1}{2n}\) et \(\frac{1}{n + 1}\), d'où \(\frac12\le u_n\le\frac{n}{n+1} < 1\). \(u_{n+1} - u_n = \frac{1}{2n + 1} + \frac{1}{2n + 2} - \frac{1}{n + 1} = \frac{1}{(2n + 1)(2n + 2)} > 0\).