Question

Bài 1: Cho b \(\in\) N, b > 1

Chứng minh: \(\dfrac{1}{b}-\dfrac{1}{b+1}< \dfrac{1}{b^2}< \dfrac{1}{b-1}-\dfrac{1}{b}\)

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

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

-Giúp tớ với, bí quá :<

Answer 1

1.

Ta có:

Vì b+1-b=1=>\(\dfrac{1}{b}-\dfrac{1}{b+1}=\dfrac{1}{b.\left(b+1\right)}\)<\(\dfrac{1}{b.b}\)(1)

Vì b-(b-1)=1=>\(\dfrac{1}{b-1}-\dfrac{1}{b}=\dfrac{1}{b.\left(b-1\right)}\)>\(\dfrac{1}{b.b}\)(2)

Từ (1) và (2)=>\(\dfrac{1}{b}-\dfrac{1}{b+1}< \dfrac{1}{b.b}< \dfrac{1}{b-1}-\dfrac{1}{b}\)

Câu 2 bạn hỏi bạn Bùi Ngọc Minh nhé PR cho nó

Answer 2

Bài 2:

Ta có:S=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{9^2}=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}\)

S>\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\left(1\right)\)

S<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}=1-\dfrac{1}{9}=\dfrac{8}{9}\left(2\right)\)

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

Answer 3

2.

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

S=\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{9.9}\)

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

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

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

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

S>\(\dfrac{2}{5}\)

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

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

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

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

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

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

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