\(\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2007^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{2006.2007}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{2006}-\dfrac{1}{2007}\)
\(=\dfrac{1}{4}-\dfrac{1}{2007}< \dfrac{1}{4}\)
\(\Rightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2007^2}< \dfrac{1}{4}\left(đpcm\right)\)
Vậy...
Đặt \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2007^2}\).
Ta thấy:
\(\dfrac{1}{5^2}< \dfrac{1}{4\cdot5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5\cdot6}\)
\(\dfrac{1}{7^2}< \dfrac{1}{6\cdot7}\)
............................
\(\dfrac{1}{2007^2}< \dfrac{1}{2006\cdot2007}\)
\(\Rightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2007^2}< \dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+...+\dfrac{1}{2006\cdot2007}\)
\(\Rightarrow A< \dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{2006}-\dfrac{1}{2007}\)
\(\Rightarrow A< \dfrac{1}{4}-\dfrac{1}{2007}\)
\(\Rightarrow A< \dfrac{1}{4}\left(đpcm\right)\)
