Sửa đề: Chứng minh A<1/2
Ta có: \(A=\frac{1}{2^2}+\frac{1}{4^2}+\cdots+\frac{1}{20^2}\)
\(=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\cdots+\frac{1}{10^2}\right)\)
Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
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\(\frac{1}{10^2}<\frac{1}{9\cdot10}=\frac19-\frac{1}{10}\)
Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}<1-\frac12+\frac12-\frac13+\cdots+\frac19-\frac{1}{10}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}<1-\frac{1}{10}<1\)
=>\(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}<1+1=2\)
=>\(A<2\cdot\frac14=\frac12\) (ĐPCM)
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