Xét tổng:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2004^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2003.2004}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}=1-\frac{1}{2004}\)
\(\Rightarrow1-A>1-\left(1-\frac{1}{2004}\right)=\frac{1}{2004}\) (đpcm)
Tại sao nó lại <\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...