Giải thích các bước giải:
Ta thấy:
$\dfrac{1}{3^2}<\dfrac{1}{2.3}$
$\dfrac{1}{4^2}<\dfrac{1}{3.4}$
$.......................$
$\dfrac{1}{99^2}<\dfrac{1}{98.99}$
$⇒\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+.....+\dfrac{1}{99^2} <\dfrac{1}{2.3}+\dfrac{1}{3.4}+......+\dfrac{1}{98.99}$$=$$\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{98}-\dfrac{1}{99}=\dfrac{1}{2}-\dfrac{1}{99}$
$⇒\dfrac{1}{2}-\dfrac{1}{99}<\dfrac{1}{2}$
$⇔\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+.....+\dfrac{1}{99^2}<\dfrac{1}{2}$
