Đặt \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2013^2}\)
\(A=\frac{1}{5\cdot5}+\frac{1}{6\cdot6}+\frac{1}{7\cdot7}+...+\frac{1}{2013\cdot2013}\)
Ta có : \(\frac{1}{5\cdot5}< \frac{1}{4\cdot5}\)
\(\frac{1}{6\cdot6}< \frac{1}{5\cdot6}\)
\(\frac{1}{7\cdot7}< \frac{1}{6\cdot7}\)
...
\(\frac{1}{2013\cdot2013}< \frac{1}{2012\cdot2013}\)
=> \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+..+\frac{1}{2013^2}< \frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{2012\cdot2013}\)
=> \(A< \frac{1}{4}-\frac{1}{2013}\)
=> \(A< \frac{2009}{8052}\)
Lại có \(\frac{2009}{8052}< \frac{1}{4}\)
Theo tính chất bắc cầu => \(A< \frac{1}{4}\)( đpcm )
Sai thì mong bạn bỏ qua