`S = 1/(2^2) + 1/(3^2) + 1/(4^2) +…+1/(9^2)`
`< 1/(1.2) + 1/(2.3) + 1/(3.4) + ... + 1/(8.9)`
`= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ..+ 1/8 - 1/9 `
`= 1 - 1/9`
`= 8/9`
`S = 1/(2^2) + 1/(3^2) + 1/(4^2) +…+1/(9^2)`
`> 1/(2.3) + 1/(3.4) + 1/(4.5) + ... + 1/(9.10)`
`= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +...+ 1/9 - 1/10 `
`= 1/2 - 1/10`
`= 2/5`
Vậy `2/5 < S < 8/9`
Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
\(...\)
\(\dfrac{1}{9^2}< \dfrac{1}{8.9}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
Mà
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(=1-\dfrac{1}{9}\)
\(=\dfrac{8}{9}\)
Lại có :
\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\)
\(=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
Mà \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}-\dfrac{1}{10}\)
\(=\dfrac{2}{5}\)
Vậy \(\dfrac{2}{5}< S< \dfrac{8}{9}\)