Chứng minh rằng :
S=[tex]\frac{1}{31}+\frac{1}{32}+\frac{1}{3}+....+\frac{1}{60}<\frac{4}{5}[/tex]
$ \dfrac{1}{31} < \dfrac{1}{30}\\\dfrac{1}{32} < \dfrac{1}{30}\\\dfrac{1}{33} < \dfrac{1}{30}\\...\\\dfrac{1}{40} < \dfrac{1}{30}\\\Rightarrow \dfrac{1}{31} + \dfrac{1}{32} + \dfrac{1}{33} + ... + \dfrac{1}{40} < 10 . \dfrac{1}{30} = \dfrac{1}{3} $
$ \dfrac{1}{41} < \dfrac{1}{40}\\\dfrac{1}{42} < \dfrac{1}{40}\\\dfrac{1}{43} < \dfrac{1}{40}\\...\\\dfrac{1}{50} < \dfrac{1}{40}\\\Rightarrow \dfrac{1}{41} + \dfrac{1}{42} + \dfrac{1}{43} + ... + \dfrac{1}{50} < 10 . \dfrac{1}{40} = \dfrac{1}{4} $
$ \dfrac{1}{51} < \dfrac{1}{50}\\\dfrac{1}{52} < \dfrac{1}{50}\\\dfrac{1}{53} < \dfrac{1}{50}\\...\\\dfrac{1}{60} < \dfrac{1}{50}\\\Rightarrow \dfrac{1}{51} + \dfrac{1}{52} + \dfrac{1}{53} + ... + \dfrac{1}{60} < 10 . \dfrac{1}{50} = \dfrac{1}{5} $
$ \Rightarrow S < \dfrac13 + \dfrac14 + \dfrac15 = \dfrac{47}{60} < \dfrac{48}{60} = \dfrac{4}{5} $