Question

[LÊ TRẦN BÁCH]

\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}\)

\(cm:\)

\(\dfrac{2}{5}< S< \dfrac{8}{9}\)

Answer 1

[SKIFGDG]

Ta có : S<\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{8.9}\)

S<\(\dfrac{2-1}{1.2}\)+\(\dfrac{3-2}{2.3}\)+...+\(\dfrac{9-8}{8.9}\)

S<1-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+...+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)

S<1-\(\dfrac{1}{9}\)

S<\(\dfrac{8}{9}\)   (1)

Lại có : S>\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{9.10}\)

S>\(\dfrac{3-2}{2.3}\)+\(\dfrac{4-3}{3.4}\)+...+\(\dfrac{10-9}{9.10}\)

S>\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)

S>\(\dfrac{1}{2}\)-\(\dfrac{1}{10}\)

S>\(\dfrac{2}{5}\)  (2)

Từ (1),(2) suy ra \(\dfrac{2}{5}\)<S<\(\dfrac{8}{9}\)