a, ta có:
[TEX] A=\frac{1}{1.2}+ \frac{1}{2.3}+ \frac{1}{3.4}+....+ \frac{1}{98.99}+ \frac{1}{99.100}=1-\frac{1}{2}+ \frac{1}{2}- \frac{1}{3}+....+ \frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}[/TEX]
b, Ta có: [TEX]B=\frac{1}{11}+\frac{1}{12}+....+\frac{1}{39}+\frac{1}{40}=(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{20})+(\frac{1}{21}+\frac{1}{22}+....+\frac{1}{30})+(\frac{1}{31}+\frac{1}{32}+....+\frac{1}{40})[/TEX]
- Có: [TEX]\frac{1}{11}+\frac{1}{12}+....+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+....+\frac{1}{20}=\frac{1}{2}[/TEX]
[TEX]\frac{1}{21}+\frac{1}{22}+....+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+....+\frac{1}{30}=\frac{1}{3}[/TEX]
[TEX]\frac{1}{31}+\frac{1}{32}+....+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+....+\frac{1}{40}=\frac{1}{4}[/TEX]
Khi đó: [TEX]B>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=1\frac{1}{2} >1>\frac{99}{100}=A[/TEX]
Vậy B>A