Đáp án + Giải thích các bước giải:
$A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{(2n)^2}$
$=\dfrac{1}{2^2}.(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2})$
$<\dfrac{1}{4}.[1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{(n-1).n}]$
$=\dfrac{1}{4}.(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n})$
$=\dfrac{1}{4}.(2-\dfrac{1}{n})$
$=\dfrac{1}{2}-\dfrac{1}{4n}$
$<\dfrac{1}{2}$
Do đó $A<\dfrac{1}{2}$
