Cho 

- Hướng dẫn giải

Đáp án cần chọn là: C

$\begin{array}{l}P=\frac{1}{2^2}+\frac{1}{3^2}+\mathrm{...}+\frac{1}{{2002}^2}+\frac{1}{{2003}^2}\\ <\frac{1}{1.2}+\frac{1}{2.3}+\mathrm{...}+\frac{1}{2001.2002}+\frac{1}{2002.2003}\\ =\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\mathrm{...}+\frac{1}{2001}\\ -\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}\\ =1-\frac{1}{2003}=\frac{2002}{2003}<1\end{array}$

Vậy P < 1