$\text{(*)}$
`A =1/1.2+1/3.4+1/5.6+...+1/99.100`
` =(1/1.2+1/3.4)+(1/5.6+...+1/99.100)`
`=7/12+(1/5.6+...+1/99.100)>7/12`
`⇒A>\frac{7}{12}`
$\text{(**)}$
`A=1-1/2+1/3-1/4+1/5-1/6+...+1/99-1/100`
`=(1+1/3+1/5+...+1/99)-(1/2+1/4+...+1/100)`
`=(1+1/2+1/3+1/4+...+1/99+1/100)-2(1/2+1/4+...+1/100)`
`=(1+1/2+1/3+...+1/100)-(1+1/2+...+1/50)`
`=1/51+1/52+...+1/100`
Dãy số trên có `50` số hạng. Vì `50` $\vdots$ `10` nên ta nhóm `10` số vào `1 ` nhóm.
`A=(1/51+1/52+...+1/60)+(1/61+1/62+...+1/70)+(1/71+1/72+...+1/80)+(1/81+...+1/90)+(1/91+...+1/100)<1/50.10+1/60.10+1/70.10+1/80.10+1/90.10=1/5+1/6+1/7+1/8+1/9<1/5+1/6+1/7.3=167/210<175/210=5/6`
`=>A<5/6`
Từ $\text{(*)(**)}$ ta cho rằng $\dfrac{7}{12}<A<\dfrac{5}{6}$.
