vì \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)(do 22 > 1.2)
\(\frac{1}{3^2}< \frac{1}{2.3}\)(do 32>2.3)
\(\frac{1}{4^2}< \frac{1}{3.4}\)(do 42 >3.4)
...
\(\frac{1}{2002^2}< \frac{1}{2001.2002}\)(do 20022 > 2001.2002)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2002^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2001.2002}\)(2)
Ta có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2001.2002}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2001}-\frac{1}{2002}\)
\(=\frac{1}{1}-\frac{1}{2002}\)
\(=\frac{2002}{2002}-\frac{1}{2002}\)
\(=\frac{2001}{2002}< 1\)(2)
Từ (1) và (2) suy ra: \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2002^2}< 1\)
Bài toán được chứng minh
