Question

Chứng minh rằng

\(\frac{1}{10^2}+\frac{1}{11^2}+\frac{1}{12^2}+...+\frac{1}{2014^2}< \frac{1}{9}\)

Answer 1

Đặt \(S=\frac{1}{10^2}+\frac{1}{11^2}+\frac{1}{12^2}+.....+\frac{1}{2014^2}\)

Ta có : \(S< \frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}+.....+\frac{1}{2013.2014}\\\)

Đặt \(A=\frac{1}{9.10}+\frac{1}{10.11}+....+\frac{1}{2013.2014}\\ =>A=\left(\frac{1}{9}-\frac{1}{10}\right)+\left(\frac{1}{10}-\frac{1}{11}\right)+......+\left(\frac{1}{2013}-\frac{1}{2014}\right)\\ =>A=\frac{1}{9}-\frac{1}{2014}\\ \)

Vậy A<\(\frac{1}{9}\)

Mà A>S =>S<\(\frac{1}{9}\)