Question

N=\(\dfrac{1}{10^2}+\dfrac{1}{11^2}+\dfrac{1}{12^2}+...+\dfrac{1}{n^2}\)

Chứng minh rằng: N<\(\dfrac{1}{9}\)

Answer 1

N = \(\dfrac{1}{10^2}+\dfrac{1}{11^2}+\dfrac{1}{12^2}+...+\dfrac{1}{n^2}\)

= \(\dfrac{1}{10.10}+\dfrac{1}{11.11}+\dfrac{1}{12.12}+...+\dfrac{1}{n.n}\)

=> N < \(\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}+...+\dfrac{1}{\left(n-1\right).n}\)

=> N < \(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(=>N< \dfrac{1}{9}-\dfrac{1}{n}\)

=> N < \(\dfrac{1}{9}\)

Vậy N < \(\dfrac{1}{9}\)