Đáp án:
Dưới
Giải thích các bước giải:
Đặt $A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}$
$⇒A=\dfrac{1}{2^2}+(\dfrac{1}{3^2}+..+\dfrac{1}{n^2})$
$⇒A=\dfrac{1}{2^2}+B$
Xét:$B$
$⇒B=(\dfrac{1}{3^2}+..+\dfrac{1}{n^2})<(\dfrac{1}{2×3}+..+\dfrac{1}{n(n+1}$
$ADCT:\dfrac{k}{n(n+k)}=\dfrac{1}{n}-\dfrac{1}{n+k}$
$⇒B<\dfrac{1}{2}-\dfrac{1}{3}+..+\dfrac{1}{n}-\dfrac{1}{n+1}$
$⇒B<\dfrac{1}{2}-\dfrac{1}{n+1}$
Do đó:$A=\dfrac{1}{2^2}+B<\dfrac{1}{2^2}+\dfrac{1}{2}-\dfrac{1}{n+1}$
$⇒A<\dfrac{3}{4}-\dfrac{1}{n+1}$
Vì $\dfrac{3}{4}-\dfrac{1}{n+1}<\dfrac{3}{4}$
$⇒A<\dfrac{3}{4}$
Vậy đpcm
