Question

Chứng minh : \(\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+...+\dfrac{1}{2005\sqrt{2004}}< 2\)

Answer 1

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

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)