Question

Bài 4. (0.5 điểm) Cho \( A = \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{8^2} + \frac{1}{9^2} \), chứng tỏ rằng \( \frac{2}{5} < A < 1 \)

Answer 1

`A= \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... + \frac{1}{8^2} + frac{1}{9^2}`

`> \frac{1}{2*3} + \frac{1]{3*4} + \frac{1}{4*5} + ... + \frac{1}{8*9} + \frac{1}{9*10}`

`= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/8 - 1/9 + 1/9 - 1/10`

`= 1/2 - 1/10`

`= 2/5`

`=> A> 2/5`      `(1)`

`A= \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... + \frac{1}{8^2} + frac{1}{9^2}`

`< \frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + ... + \frac{1}{7*8} + frac{1}{8*9}`

`= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/7 - 1/8 + 1/8 - 1/9`

`= 1 - 1/9 < 1`

`=> A<1`      `(2)`

Từ `(1),(2)=>2/5<A<1`   `(đpcm)`