Ta có : \(\dfrac{1}{\left(n+1\right)\sqrt{n}}=\dfrac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\sqrt{n}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\left(\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}\right)=\left(1+\dfrac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)< 2\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
Áp dụng điều này vào bài toán , ta có :
\(\left\{{}\begin{matrix}\dfrac{1}{2\sqrt{1}}< 2\left(1-\dfrac{1}{\sqrt{2}}\right)=2-\sqrt{2}\\\dfrac{1}{3\sqrt{2}}< 2\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\right)=\sqrt{2}-\dfrac{2}{\sqrt{3}}\\....\\\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)=\dfrac{2}{\sqrt{n}}-\dfrac{2}{\sqrt{n+1}}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{2\sqrt{1}}+\dfrac{1}{2\sqrt{3}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2-\dfrac{2}{\sqrt{n+1}}< 2\) ( Sửa đề ^-^ )
