$A = \dfrac{1}{1^2}+\dfrac{1}{2^2}+....+\dfrac{1}{100^2}$
$ < \dfrac{1}{1} + \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{99.100}$
$ = 1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{99}-\dfrac{1}{100}$
$ = 1+1-\dfrac{1}{100} = \dfrac{199}{100}$
Vậy $A < \dfrac{199}{100}$