Giả sử: $A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}$
Ta có:
$\dfrac{1}{2^2}<\dfrac{1}{1.2}$
.......
$\dfrac{1}{100^2}<\dfrac{1}{99.100}$
$⇒A<\dfrac{1}{1.2}+....+\dfrac{1}{99.100}$
$⇒A<1-\dfrac{1}{2}+....+\dfrac{1}{99}-\dfrac{1}{100}$
$⇒A<1-\dfrac{1}{100}$
$⇒A<\dfrac{99}{100}$
mà $\dfrac{99}{100}<1$
$⇒A<1$
Vậy $\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}<1$
