Sửa đề: \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{2008^2}\)
Ta có: \(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
\(\frac{1}{4^2}<\frac{1}{3\cdot4}=\frac13-\frac14\)
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\(\frac{1}{2008^2}<\frac{1}{2007\cdot2008}=\frac{1}{2007}-\frac{1}{2008}\)
Do đó: \(\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{2008^2}<\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{2007}-\frac{1}{2008}=\frac12-\frac{1}{2008}<\frac12\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2008^2}<\frac14+\frac12\)
=>\(A<\frac14+\frac24=\frac34\)
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