Ta có: A = $\frac{1}{2^2}$ + $\frac{1}{3^2}$ + ... + $\frac{1}{2019^2}$
Ta thấy:
$\frac{1}{2^2}$ < $\frac{1}{1.2}$
$\frac{1}{3^2}$ < $\frac{1}{2.3}$
........................
$\frac{1}{2019^2}$ < $\frac{1}{2018.2019}$
=> A < $\frac{1}{1.2}$ + $\frac{1}{2.3}$ + .... + $\frac{1}{2018.2019}$
<=> A < 1 - $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ +...+ $\frac{1}{2018}$ - $\frac{1}{2019}$
<=> A < 1 - $\frac{1}{2019}$ = $\frac{2018}{2019}$ > $\frac{3}{4}$
Vậy A > $\frac{3}{4}$