Cách giải:Chứng minh B>0(Như đã nói ở trên)
$B=1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-\dfrac{1}{4^2}-......-\dfrac{1}{2004^2}$
$\to B=1-\underbrace{(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+......+\dfrac{1}{2004^2})}_{A}$
$A<\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+......+\dfrac{1}{2003.2004}$
$\to A<1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+......+\dfrac{1}{2003}-\dfrac{1}{2004}$
$\to A<1-\dfrac{1}{2004}$
$\to B>1-(1-\dfrac{1}{2004})$
$\to B>\dfrac{1}{2004}>0$