Nếu :$A=\dfrac{1}{1.2} + \dfrac{3}{2.5} + \dfrac{9}{5.14} + \dfrac{23}{14.37} + \dfrac{15}{37.52} + \dfrac{1967}{52.2019}$
$⇔ A = 1 - \dfrac{1}{2} +\dfrac{1}{2} - \dfrac{1}{5} + .......... + \dfrac{1}{52} - \dfrac{1}{2019}$
$⇔ A =1 - \dfrac{1}{2019}$
Vì $1 - \dfrac{1}{2019}$ < $1$ $⇒$ $A<1$
Vậy $A<1$($đpcm$)
Nếu:$A=\dfrac{1}{1.2} + \dfrac{3}{2.5} + \dfrac{9}{5.14} + \dfrac{23}{14.37} + \dfrac{15}{37.52} + \dfrac{2567}{52.2619}$
$⇔ A = 1 - \dfrac{1}{2} +\dfrac{1}{2} - \dfrac{1}{5} + .......... + \dfrac{1}{52} - \dfrac{1}{2619}$
$⇔ A =1 - \dfrac{1}{2619}$
Vì $1 - \dfrac{1}{2169}$ < $1$ $⇒$ $A<1$
Vậy $A<1$($đpcm$)
