Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)và B= \(\frac{8}{9}\)
Ta có :\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)
\(........\)
\(\frac{1}{9^2}< \frac{1}{8\cdot9}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{9^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{8\cdot9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{9^2}=\frac{8}{9}\)
Vậy \(A=B\)
