Giải thích các bước giải:
Ta có :
$A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+..+\dfrac{1}{(2n)^2}$
$\to A=\dfrac{1}{2^2}(1+\dfrac{1}{2^2}+..+\dfrac{1}{n^2})$
$\to A<\dfrac{1}{2^2}(1+\dfrac{1}{1.2}+..+\dfrac{1}{(n-1)n})$
$\to A<\dfrac{1}{2^2}(1+1-\dfrac{1}2+\dfrac{1}2-\dfrac{1}3+..+\dfrac{1}{n-1}-\dfrac{1}n)$
$\to A<\dfrac{1}{4}(2-\dfrac{1}n)$
$\to A<\dfrac{1}4.2=\dfrac{1}2$
