Question

[AM]

Cho S =  \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ \(\frac{1}{4^2}\)+ . . . . . . + \(\frac{1}{9^2}\)

Chứng tỏ rằng \(\frac{2}{5}\) <  S  <   \(\frac{8}{9}\)

Answer 1

[Chim Hoạ Mi]

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

\(S>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

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

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

\(S>\frac{4}{10}=\frac{2}{5}\)

Answer 2

[hà phương uyên]

\(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{9.10}< S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3\cdot4}+...+\frac{1}{8.9}\)

=>\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}< S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{8}-\frac{1}{9}\)

=> \(\frac{1}{2}-\frac{1}{10}< S< 1-\frac{1}{9}\)

=> \(\frac{2}{5}< S< \frac{8}{9}\)(dpcm )