\(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}\)
\(S>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(S>\frac{1}{2}-\frac{1}{10}\)
\(S>\frac{4}{10}=\frac{2}{5}\)
\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9.10}< S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3\cdot4}+...+\frac{1}{8.9}\)
=>\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}< S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{8}-\frac{1}{9}\)
=> \(\frac{1}{2}-\frac{1}{10}< S< 1-\frac{1}{9}\)
=> \(\frac{2}{5}< S< \frac{8}{9}\)(dpcm )
ta có \(\frac{1}{2\times3}\)< \(\frac{1}{2^2}\)\(=\)\(\frac{1}{2\times2}\)<\(\frac{1}{1\times2}\)
\(\frac{1}{3\times4}\)<\(\frac{1}{3^2}\)\(=\)\(\frac{1}{3\times3}\)<\(\frac{1}{2\times3}\)
.......................................................................
\(\frac{1}{9\times10}\)<\(\frac{1}{9^2}\)\(=\)\(\frac{1}{9\times9}\)<\(\frac{1}{8\times9}\)
\(\Rightarrow\)\(\frac{1}{2\times3}\)+\(\frac{1}{3\times4}\)+.............+\(\frac{1}{9\times10}\)<S<\(\frac{1}{1\times2}\)+\(\frac{1}{2\times3}\)+..........+\(\frac{1}{8\times9}\)
\(\Rightarrow\)\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}\text{+}........\text{+}\frac{1}{9}-\frac{1}{10}\)<S<\(\frac{1}{1}-\frac{1}{2}\text{+}\frac{1}{2}-\frac{1}{3}\text{+}...........\text{+}\frac{1}{8}-\frac{1}{9}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{10}< S< \frac{1}{1}-\frac{1}{9}\)
\(\Rightarrow\frac{2}{5}< S< \frac{8}{9}\)
Vậy \(\frac{2}{5}< S< \frac{8}{9}\)
