câu 22:
Ta có: [tex]\frac{1}{k\sqrt{k+1} +(k+1)\sqrt{k}}=\frac{1}{\sqrt{k(k+1)}}.\frac{1}{\sqrt{k} +\sqrt{k+1}}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k(k+1)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}[/tex]
=> un= [tex]\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}=1-\frac{1}{\sqrt{n+1}}[/tex]
=> lim un=1