Question

[Chu Khánh Linh]

1. Cho A=1/101+1/102+...+1/200

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Chứng minh 1/2<A<1

Answer 1

[Mây]

Ta có : \(\frac{1}{101}>\frac{1}{200};\frac{1}{102}>\frac{1}{200};...;\frac{1}{199}>\frac{1}{200}\)

=> \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{199}+\frac{1}{200}>\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}=\frac{100}{200}=\frac{1}{2}\)

=> \(A>\frac{1}{2}\)   (1) 

\(\frac{1}{101}<\frac{1}{100};\frac{1}{102}<\frac{1}{100};...;\frac{1}{199}<\frac{1}{100};\frac{1}{200}<\frac{1}{100}\)

=> \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{199}+\frac{1}{200}<\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{100}{100}=1\)

=> \(A<1\)    (2)

Từ (1) và (2) => 1/2 < A <1