Question

chứng tỏ \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}< 1\)

giúp mình nhanh nh

Answer 1

ta có:

\(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}\)

\(\Rightarrow B< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)

\(\Rightarrow B< \frac{1}{1}-\frac{1}{9}< 1\)

vậy B < 1

Answer 2

Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{7\cdot8}\)

\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)

...

\(\frac{1}{8^2}< \frac{1}{7\cdot8}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)

\(A=1-\frac{1}{8}< 1\)

\(B< A< 1\left(đpcm\right)\)