Question

Chứng minh : \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)

Answer 1

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

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

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

=> biểu thức trên sẽ \(< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+......+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

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

Answer 2

Ta có: \(\frac{1}{\left(n+1\right)\sqrt{n}}\)

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

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

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

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

=> Biểu thức trên sẽ bé hơn \(2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

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