\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)
Biểu thức có 200 số hạng
Ta có: \(\frac{1}{101}>\frac{1}{300};\frac{1}{102}>\frac{1}{300};...;\frac{1}{299}>\frac{1}{300};\frac{1}{300}=\frac{1}{300}\)
\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}>\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}=\frac{200}{300}=\frac{2}{3}\)
Vậy....
Ta có : \(\frac{1}{101}>\frac{1}{300}\)
\(\frac{1}{102}>\frac{1}{300}\)
..................
\(\frac{1}{300}=\frac{1}{300}\)
Do đó \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{299}+\frac{1}{300}>\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\)
Hay \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}>200\cdot\frac{1}{300}=\frac{2}{3}\Rightarrowđpcm\)
