Đáp án: $A<1$
Giải thích các bước giải:
Ta có :
$A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+..+\dfrac{1}{n^2}$
$\to A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+..+\dfrac{1}{n.n}$
$\to A<\dfrac{1}{1.2}+\dfrac{1}{2.3}+..+\dfrac{1}{(n-1).n}$
$\to A<\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+..+\dfrac{n-(n-1)}{(n-1).n}$
$\to A<\dfrac11-\dfrac12+\dfrac12-\dfrac13+...+\dfrac{1}{n-1}-\dfrac1n$
$\to A<1-\dfrac1n<1$
