Question

[My Phan]

Cho các số:\(a_1,a_2,a_3,...,a_{2009}\) được xác định theo công thức sau:

\(a_n=\frac{2}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}\) với n=1,2,3,...,2008

Chứng minh rằng :\(a_1+a_2+a_3+...+a_{2009< \frac{2008}{2010}}\)

Answer 1

[Phùng Minh Quân]

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

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

\(a_1+a_2+a_3+...+a_{2009}< 1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...-\frac{1}{\sqrt{2010}}=1-\frac{1}{\sqrt{2010}}< \frac{2008}{2010}\)