Cho tổng

- Hướng dẫn giải

$A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\mathrm{...}+\frac{1}{99}+\frac{1}{100}$$=\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\mathrm{...}+\frac{1}{90}\right)+\left(\frac{1}{91}+\frac{1}{92}+\mathrm{...}\frac{1}{99}+\frac{1}{100}\right)$

Đặt $A_1=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\mathrm{...}+\frac{1}{90}$; $A_2=\frac{1}{91}+\frac{1}{92}+\mathrm{...}\frac{1}{99}+\frac{1}{100}$

Ta có $\frac{1}{10}>\frac{1}{90};\frac{1}{11}>\frac{1}{90};\frac{1}{12}>\frac{1}{90};\mathrm{...};\frac{1}{89}>\frac{1}{90}$

$A_1=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\mathrm{...}+\frac{1}{90}>81\cdot \frac{1}{90}$

$A_1=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\mathrm{...}+\frac{1}{90}>81\cdot \frac{1}{90}$

Tương tự $\frac{1}{91}>\frac{1}{100};\frac{1}{92}>\frac{1}{100};\mathrm{...};\frac{1}{99}>\frac{1}{100}$

$A_2=\frac{1}{91}+\frac{1}{92}+\mathrm{...}\frac{1}{99}+\frac{1}{100}>10\cdot \frac{1}{100}$

$A_2=\frac{1}{91}+\frac{1}{92}+\mathrm{...}\frac{1}{99}+\frac{1}{100}>\frac{10}{100}=\frac{1}{10}$

$A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\mathrm{...}+\frac{1}{99}+\frac{1}{100}=A_1+A_2>\frac{9}{10}+\frac{1}{10}$

$A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\mathrm{...}+\frac{1}{99}+\frac{1}{100}>1$