Gọi biểu thức bên VT là A
Ta có:
[tex]\frac{1}{(n+1)\sqrt{n}}=\frac{\sqrt{n}}{n(n+1)}=\sqrt{n}(\frac{1}{n}-\frac{1}{n+1})=\sqrt{n}(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}})(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})\\ =(1+\frac{\sqrt{n}}{\sqrt{n+1}})(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})<2(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})[/tex]
Thay vào biểu thức ta có:
[tex]A=\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+....+\frac{1}{(n+1)\sqrt{n}}\\ A< 2(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}})\\ A<2(1-\frac{1}{\sqrt{n+1}})< 2[/tex]