Đề bài : Cho \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}\).
Chứng minh : \(\frac{8}{33}< A< \frac{2}{5}\).
Giải : Ta có : \(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}< A< \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}\)
\(\frac{1}{3}-\frac{1}{11}< A< \frac{1}{2}-\frac{1}{10}\)
\(\frac{1}{11}-\frac{3}{33}=\frac{8}{22}< A< \frac{5}{10}-\frac{1}{10}=\frac{2}{5}\)
\(\frac{8}{33}< A< \frac{2}{5}\)
Ta có: \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+....+\frac{1}{10^2}< \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9\cdot10}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow A< \frac{2}{5}\)(1)
Lại có: \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}>\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}\)
\(\Rightarrow A>\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow A>\frac{8}{33}\)(2)
Từ (1)(2) suy ra \(\frac{8}{33}< A< \frac{2}{5}\)
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