\(\frac{1}{(n+1)\sqrt{n} }=\frac{\sqrt{n} }{n(n+1)}=\sqrt{n} (\frac{1}{\sqrt{n} } -\frac{1}{\sqrt{n+1} } )(\frac{1}{\sqrt{n} } +\frac{1}{\sqrt{n+1} } )=(1+\frac{\sqrt{n} }{\sqrt{n+1} } )(\frac{1}{\sqrt{n} } -\frac{1}{\sqrt{n+1} } <2(\frac{1}{\sqrt{n} } -\frac{1}{\sqrt{n+1} } )\)

Áp dụng BĐT vừa CM ta có

A< 2(1-\(\frac{1}{\sqrt{2} } +\frac{1}{\sqrt{2} } -\frac{1}{\sqrt{3} } +...+\frac{1}{\sqrt{n} } -\frac{1}{\sqrt{n+1} } \))<2(đpcm)

Xét dạng tổng quát:

\(\frac{1}{\left(n+1\right)\sqrt{n}}=\frac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}.\frac{1}{n\left(n+1\right)}=\sqrt{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)

\(=\sqrt{n}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\right)=\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(1+\frac{\sqrt{n}}{\sqrt{n+1}}\right)\)

\(< \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\left(1+\frac{\sqrt{n+1}}{\sqrt{n+1}}\right)=2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Áp dụng vào bài toán:

\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2005\sqrt{2004}}\)

\(< 2\left(1-\frac{1}{\sqrt{2}}\right)+2\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right)+...+\left(\frac{1}{\sqrt{2003}}-\frac{1}{\sqrt{2004}}\right)\)

\(< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{2003}}-\frac{1}{\sqrt{2004}}\right)\)

\(< 2\left(1-\frac{1}{\sqrt{2004}}\right)\)

\(< 2-\frac{2}{\sqrt{2004}}< 2\)

=>đpcm

Ta có:

\(\dfrac{1}{\sqrt{n}}=\dfrac{2}{2\sqrt{n}}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{n}}< \dfrac{2}{\sqrt{n-1}+\sqrt{n}}=2\left(\sqrt{n}-\sqrt{n-1}\right)\\\dfrac{1}{\sqrt{n}}>\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=2\left(\sqrt{n+1}-\sqrt{n}\right)\end{matrix}\right.\)

Thế vô giải tiếp

\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}...+\dfrac{1}{\sqrt{79}+\sqrt{80}}\)

\(=\left(\sqrt{2}-\sqrt{1}\right)+\left(\sqrt{3}-\sqrt{2}\right)+...+\left(\sqrt{80}-\sqrt{79}\right)\)

\(=\sqrt{80}-\sqrt{2}\)

Đến đây bấm máy rồi đối chiếu kết quả cho nhanh, hoặc nếu em thik "màu mè" hơn thì giả sử lớn hơn rồi biến đổi tương đương thôi :)

\(\dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\dfrac{\sqrt{k}-\sqrt{k+1}}{k-k-1}=\sqrt{k+1}-\sqrt{k}\\ \Leftrightarrow\text{Đặt}\text{ }A=\dfrac{1}{3\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{2\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{2\left(\sqrt{2011}+\sqrt{2010}\right)}\\ \Leftrightarrow A< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{2011}+\sqrt{2010}}\right)\)

\(\Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2011}-\sqrt{2010}\right)\\ \Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2011}-1\right)< \dfrac{1}{2}\cdot\dfrac{\sqrt{2011}-1}{\sqrt{2011}}=\dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)