Question

[Thầy Đức Anh]

Cho $S=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+\ldots+\dfrac{1}{60}$.

Chứng minh rằng: $\dfrac{3}{5}<S<\dfrac{4}{5}$.

Answer 1

[Trần Nguyễn Khánh Hân]

+ $S=\genfrac{}{}{0ex}{}{1}{31}+\genfrac{}{}{0ex}{}{1}{32}+\genfrac{}{}{0ex}{}{1}{33}+\dots +\genfrac{}{}{0ex}{}{1}{60}$

$S<\left(\genfrac{}{}{0ex}{}{1}{30}+\genfrac{}{}{0ex}{}{1}{30}+\dots +\genfrac{}{}{0ex}{}{1}{30}\right)+\left(\genfrac{}{}{0ex}{}{1}{40}+\genfrac{}{}{0ex}{}{1}{40}+\dots +\genfrac{}{}{0ex}{}{1}{40}\right)+\left(\genfrac{}{}{0ex}{}{1}{50}+\genfrac{}{}{0ex}{}{1}{50}+\dots +\genfrac{}{}{0ex}{}{1}{50}\right)$

$S<\genfrac{}{}{0ex}{}{10}{30}+\genfrac{}{}{0ex}{}{10}{40}+\genfrac{}{}{0ex}{}{10}{50}<\genfrac{}{}{0ex}{}{48}{60}=\genfrac{}{}{0ex}{}{4}{5};$

+ $S>\left(\genfrac{}{}{0ex}{}{1}{40}+\genfrac{}{}{0ex}{}{1}{40}+\dots +\genfrac{}{}{0ex}{}{1}{40}\right)+\left(\genfrac{}{}{0ex}{}{1}{50}+\genfrac{}{}{0ex}{}{1}{50}+\dots +\genfrac{}{}{0ex}{}{1}{50}\right)+\left(\genfrac{}{}{0ex}{}{1}{60}+\genfrac{}{}{0ex}{}{1}{60}+\dots +\genfrac{}{}{0ex}{}{1}{60}\right)$

$S>\genfrac{}{}{0ex}{}{10}{40}+\genfrac{}{}{0ex}{}{10}{50}+\genfrac{}{}{0ex}{}{10}{60}>\genfrac{}{}{0ex}{}{3}{5}.$