Ta có:
\(\frac{1}{\left(k+1\right)\sqrt{k}}=\sqrt{k}\left(\frac{1}{k\left(k+1\right)}\right)=\sqrt{k}\left(\frac{1}{k}-\frac{1}{k+1}\right)\)
\(=\sqrt{k}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\left(\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}\right)\)
\(=\sqrt{k}\left(\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}\right)\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)
\(=\left(1+\frac{\sqrt{k}}{\sqrt{k+1}}\right)\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)< 2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)
Từ điều này ta suy ra:
\(A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=>A< 2\left(1-\frac{1}{\sqrt{n+1}}\right)< 2\)
