Chứng minh rằng: 1/101 + 1/102+ 1/103+...+1/150>1/3

* Hướng dẫn giải

$a)\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+\mathrm{...}+\frac{1}{149}+\frac{1}{150}>\underset{50\text{s}o}{\underbrace{\frac{1}{150}+\frac{1}{150}+\mathrm{...}+\frac{1}{150}}}=\frac{1}{3}$

$b)\frac{1}{201}+\frac{1}{202}+\mathrm{...}+\frac{1}{400}<\underset{200\text{s}o}{\underbrace{\frac{1}{200}+\frac{1}{200}+\mathrm{...}+\frac{1}{200}}}=1$

c) đặt S=$\left(\frac{1}{31}+\mathrm{...}+\frac{1}{40}\right)+\left(\frac{1}{41}+\mathrm{...}+\frac{1}{50}\right)+\left(\frac{1}{51}+\mathrm{...}+\frac{1}{60}\right)$

$\begin{array}{l}S>\underset{10\text{s}o}{\underbrace{\frac{1}{40}+\mathrm{..}+\frac{1}{40}}}+\underset{10\text{s}o}{\underbrace{\frac{1}{50}+\mathrm{...}+\frac{1}{50}}}+\underset{10\text{s}o}{\underbrace{\frac{1}{60}+\mathrm{...}+\frac{1}{60}}}\\ \\ =\frac{1}{40}+\frac{1}{50}+\frac{1}{60}=\frac{37}{60}\end{array}$

$Mà \frac{37}{60}>\frac{36}{60}=\frac{3}{5}=>S>\frac{3}{5}$

$\begin{array}{l}S<\underset{10\text{s}o}{\underbrace{\frac{1}{30}+\mathrm{..}+\frac{1}{30}}}+\underset{10\text{s}o}{\underbrace{\frac{1}{40}+\mathrm{...}+\frac{1}{40}}}+\underset{10\text{s}o}{\underbrace{\frac{1}{50}+\mathrm{...}+\frac{1}{50}}}\\ \\ =\frac{10}{30}+\frac{10}{40}+\frac{10}{50}=\frac{47}{60}\end{array}$

$Mà \frac{47}{60}<\frac{48}{60}=\frac{4}{5}=>S<\frac{4}{5}$