Giải thích các bước giải:
Ta có :
$A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac1{3^2}+..+\dfrac{1}{2018^2}$
$\to A=\dfrac{1}{1.1}+\dfrac{1}{2.2}+\dfrac{1}{3.3}+..+\dfrac{1}{2018.2018}$
$\to A<\dfrac{1}{1.1}+\dfrac{1}{1.2}+\dfrac{1}{2.3}+..+\dfrac{1}{2017.2018}$
$\to A<1+\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+..+\dfrac{2018-2017}{2017.2018}$
$\to A<1+\dfrac11-\dfrac12+\dfrac12-\dfrac13+...+\dfrac1{2017}-\dfrac1{2018}$
$\to A<1+1-\dfrac1{2018}$
$\to A<2-\dfrac{1}{2018}$
$\to A<2-\dfrac14$
$\to A<\dfrac74$
