`A=1/101+1/102+1/103+...+1/109+1/200`

`=(1/101+1/102+...+1/125)+(1/126+1/127+...+1/150)+(1/151+1/152+...+1/175)+(1/176+1/177+...+1/200)`

`>\underbrace{1/125+1/125+...+1/125}_{25\text{số}}+\underbrace{1/150+1/150+1/150+...+1/150}_{25\text{số}}+\underbrace{1/175+1/175+...+1/175}_{25\text{số}}+\underbrace{1/200+1/200+...+1/200}_{25\text{số}}`

`=25/125+25/150+25/175+25/200=1/5+1/6+1/7+1/8`

`=12/35+1/6+1/8>1/3+1/6+1/8=1/2+1/8=5/8>13/21`

 Vậy `A>13/21`