Ta có
\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)=\(\sqrt{n}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
\(=\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\left(1-\frac{1}{\sqrt{n+1}}\right)< 2\left(\frac{1}{n}-\frac{1}{\sqrt{n+1}}\right)\)
nên \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+.....+\frac{1}{\left(n+1\right)\sqrt{n}}\)\(< 2\left(\left(\frac{1}{n}-\frac{1}{\sqrt{n+1}}\right)+...+\left(3\sqrt{2}-2\right)+\left(2-1\right)\right)\) = 2
