Question

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

Answer 1

Chứng minh biểu thức đó <2

Với mọi \(n\in N^{\cdot}\), ta có

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

\(\Leftrightarrow1< 2\left(n+1\right).\sqrt{n}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)

\(\Leftrightarrow0< n+1-2\sqrt{n+1}.\sqrt{n}+n\)

\(\Leftrightarrow0< \left(\sqrt{n+1}-\sqrt{n}\right)^2\)(Luôn đúng vì n thuộc N^*)

Do đó: \(\dfrac{1}{2}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+...\dfrac{1}{2005\sqrt{2004}}< 2\left(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2004}}-\dfrac{1}{\sqrt{2005}}\right)\)

\(=2\left(1-\dfrac{1}{\sqrt{2005}}\right)< 2\)