Ta có :
\(\dfrac{1}{5^2}< \dfrac{1}{4.5}\)
\(\dfrac{1}{6^2}< \dfrac{1}{5.6}\)
....................
\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Leftrightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+........+\dfrac{1}{100^2}< \dfrac{1}{4.5}+\dfrac{1}{5.6}+.....+\dfrac{1}{99.100}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{4}-\dfrac{1}{100}< \dfrac{1}{4}\left(1\right)\)
Lại có :
\(\dfrac{1}{5^2}>\dfrac{1}{5.6}\)
\(\dfrac{1}{6^2}>\dfrac{1}{6.7}\)
............
\(\dfrac{1}{100^2}>\dfrac{1}{100.101}\)
\(\Leftrightarrow\dfrac{1}{5^2}+\dfrac{1}{6^2}+......+\dfrac{1}{100^2}>\dfrac{1}{5.6}+\dfrac{1}{6.7}+.....+\dfrac{1}{100.101}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+.....+\dfrac{1}{100}-\dfrac{1}{101}=\dfrac{1}{5}-\dfrac{1}{101}>\dfrac{1}{6}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
