Réponse :
Bonjour,
Explications étape par étape
[tex]u_n=\sum_{k=n}^{2n}\ \dfrac{1}{k} \\\\u_{n+1}=\sum_{k=n+1}^{2n+2}\ \dfrac{1}{k} \\\\u_{n+1}-u_n=\sum_{k=n+1}^{2n+2}\ \dfrac{1}{k} -\sum_{k=n}^{2n}\ \dfrac{1}{k}\\\\=\sum_{k=n+1}^{2n}\ \dfrac{1}{k}+\sum_{k=2n+1}^{2n+2}\ \dfrac{1}{k}-(\sum_{k=n}^{n}\ \dfrac{1}{k}+\sum_{k=n+1}^{2n}\ \dfrac{1}{k})\\\\=\sum_{k=2n+1}^{2n+2}\ \dfrac{1}{k}-\sum_{k=n}^{n}\ \dfrac{1}{k}\\\\=\dfrac{1}{2n+1} +\dfrac{1}{2n+2} -\dfrac{1}{n} \\\\=\dfrac{n*(2n+2)+n*(2n+1)-(2n+1)(2n+2)}{2n(n+1)(2n+1)} \\[/tex]
[tex]=\dfrac{2n^2+2n+2n^2+n-4n^2-6n-2}{2n(n+1)(2n+1)} \\\\=\dfrac{-3n-2}{2n(n+1)(2n+1)} \\[/tex]
