Lời giải:
Ta có:
\((\sqrt{n+1}+\sqrt{n})U_n=\frac{2}{2n+1}\)
\(\Rightarrow U_n=\frac{2}{(2n+1)(\sqrt{n+1}+\sqrt{n})}=\frac{2(\sqrt{n+1}-\sqrt{n})}{2n+1}\)
\(=\frac{2(\sqrt{n+1}-\sqrt{n})}{(n+1)+n}<\frac{2(\sqrt{n+1}-\sqrt{n})}{2\sqrt{n(n+1)}}\) (áp dụng bđt am-gm thì \((n+1)+n\geq 2\sqrt{n(n+1)}\), dấu bằng không xảy ra vì \(n\neq n+1\))
hay \(U_n< \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Do đó:
\(U_1+U_2+...+U_{2010}< \frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-...+\frac{1}{\sqrt{2010}}-\frac{1}{\sqrt{2011}}\)
\(\Leftrightarrow U_1+U_2+..+U_{2010}< 1-\frac{1}{\sqrt{2011}}< \frac{1005}{1006}\)
Ta có đpcm.
