Câu trả lời:
Ta có: \(\frac{1}{n.\left(n+1\right)}<\frac{1}{n^2}<\frac{1}{\left(n-1\right).n}\) (n>1, n dương)
=> \(\frac{1}{n}-\frac{1}{n+1}<\frac{1}{n^2}<\frac{1}{n-1}-\frac{1}{n}\)
=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\) < A < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}\)
=> \(\frac{2}{5}\) < A < \(\frac{8}{9}\)