Cho P= 1/ 2^2+ 1/ 3^2+....+ 1/ 2002^2+ 1/ 2003^2. Chọn câu đúng

* Hướng dẫn giải

$P=\frac{1}{2^2}+\frac{1}{3^2}+\mathrm{...}+\frac{1}{{2002}^2}+\frac{1}{{2003}^2}$

$<\frac{1}{1.2}+\frac{1}{2.3}+\mathrm{...}+\frac{1}{2001.2002}+\frac{1}{2002.2003}$

$=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\mathrm{...}+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}$

$=1-\frac{1}{2003}=\frac{2002}{2003}<1$