- Đặt \(A=1-\frac{1}{2^2}-\frac{1}{3^2}-...-\frac{1}{2006^2}\)
- Ta có: \(1=1\)
\(\frac{1}{2^2}>\frac{1}{2.3}\)
\(\frac{1}{3^2}>\frac{1}{3.4}\)
\(................\)
\(\frac{1}{2006^2}>\frac{1}{2006.2007}\)
\(\Rightarrow A>1-\frac{1}{2.3}-\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{2006.2007}\)
\(\Leftrightarrow A>1-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)-...-\left(\frac{1}{2006}-\frac{1}{2007}\right)\)
\(\Leftrightarrow A>1-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+\frac{1}{4}-...-\frac{1}{2006}+\frac{1}{2007}\)
\(\Leftrightarrow A>1+\frac{1}{2007}=\frac{2008}{2007}\)mà \(\frac{2008}{2007}>1>\frac{1}{2006}\)
\(\Rightarrow A>\frac{1}{2006} \left(ĐPCM\right)\)
^_^ Chúc bạn hok tốt ^_^
