Réponse : Bonjour,
[tex]\displaystyle u_{n+1}-u_{n}=\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n+2}-(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n})\\=-\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}=\frac{-2}{2n+2}+\frac{1}{2n+1}+\frac{1}{2n+2}=-\frac{1}{2n+2}+\frac{1}{2n+1}\\=\frac{-(2n+1)+2n+2}{2(n+1)(2n+1)}=\frac{-2n-1+2n+2}{2(n+1)(2n+1)}=\frac{1}{2(n+1)(2n+1)}[/tex]
