Question

[Nguyễn Thị Hiền Nga]

1.Chứng minh rằng :\(\dfrac{5}{8}< \dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}< \dfrac{3}{4}\)

Answer 1

[Phạm Nguyễn Tất Đạt]

+)Đặt \(A=\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}\)

\(A=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{125}\right)+\left(\dfrac{1}{126}+\dfrac{1}{127}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+...+\dfrac{1}{175}\right)+\left(\dfrac{1}{176}+...+\dfrac{1}{200}\right)\)\(A>\dfrac{1}{125}.25+\dfrac{1}{150}.25+\dfrac{1}{175}.25+\dfrac{1}{200}.25=\dfrac{533}{840}>\dfrac{5}{8}\)

+)\(A=\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}\)

\(A=\left(\dfrac{1}{101}+...+\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+...+\dfrac{1}{140}\right)+\left(\dfrac{1}{141}+...+\dfrac{1}{160}\right)+\left(\dfrac{1}{161}+...+\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+...+\dfrac{1}{200}\right)\)\(A< \dfrac{1}{100}.20+\dfrac{1}{120}.20+\dfrac{1}{140}.20+\dfrac{1}{160}.20+\dfrac{1}{180}.20=\dfrac{1879}{2520}< \dfrac{3}{4}\)