Question

Chứng minh rằng : \(\frac{1}{6}\)< \(\frac{1}{5^2}\)+\(\frac{1}{6^2}\)+...+\(\frac{1}{100^2}\)<\(\frac{1}{4}\)

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Answer 1

 Ta xét A= \(\frac{1}{5^2}+\frac{1}{6^2}+..+\frac{1}{100^2}\)

\(\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}...+\frac{1}{100.101}\)

=> \(A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)

=> \(A>\frac{1}{5}-\frac{1}{101}\)

=> \(A>\frac{96}{505}>\frac{96}{576}=\frac{1}{4}\)

Ta có : \(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

=> \(A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

=> \(A< \frac{1}{4}-\frac{1}{100}\)

=> \(A< \frac{6}{25}< \frac{6}{24}=\frac{1}{4}\)

Answer 2

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