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Ta có: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{2^2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)\(=\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{9}=\dfrac{23}{36}< \dfrac{32}{36}=\dfrac{8}{9}\). (1)
Ta lại có: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2^2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{10}=\dfrac{19}{20}>\dfrac{8}{20}=\dfrac{2}{5}\). (2)
Từ (1) và (2) suy ra đpcm.
a)Ta có:\(\dfrac{1}{b}-\dfrac{1}{b+1}=\dfrac{b+1-b}{b\left(b+1\right)}=\dfrac{1}{b^2+b}< \dfrac{1}{b^2}\)(do b>1)
\(\dfrac{1}{b-1}-\dfrac{1}{b}=\dfrac{b-b+1}{\left(b-1\right)b}=\dfrac{1}{b^2-b}>\dfrac{1}{b^2}\)(do b>1)
b)Áp dụng từ câu a
=>\(\dfrac{1}{2}-\dfrac{1}{3}< \dfrac{1}{2^2}< \dfrac{1}{1}-\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{4}< \dfrac{1}{3^2}< \dfrac{1}{2}-\dfrac{1}{3}\)
.........................
\(\dfrac{1}{9}-\dfrac{1}{10}< \dfrac{1}{9^2}< \dfrac{1}{8}-\dfrac{1}{9}\)
=>\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}< S< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
=>\(\dfrac{1}{2}-\dfrac{1}{10}< S< 1-\dfrac{1}{9}\)
=>\(\dfrac{2}{5}< S< \dfrac{8}{9}\)(đpcm)
\(S>\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{9.10}\)
\(S>\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\) (1)
\(S< \dfrac{1}{1.2}+\dfrac{1}{2.3}+....+\dfrac{1}{8.9}\)
\(S< 1-\dfrac{1}{9}=\dfrac{8}{9}\) (2)
(1) và (2) => đpcm
Ta có:
1/2^2 > 1/2.3
1/3^2 > 1/3.4
...
1/10^2 > 1/10.11
-> Cộng dọc theo vế ta có:
1/2^2+1/3^2+...+1/10^2 > 1/2.3+1/3.4+...+1/10.11
= 1/2-1/3+1/3-1/4+...+1/10-1/11
= 1/2 - 1/11 = 9/22 (đpcm)
1/31>1/40
1/32>1/40
...
1/40=1/40
=>1/31+1/32+...+1/40>1/40*10=1/4
1/41>1/50
1/42>1/50
...
1/50=1/50
=>1/41+1/42+...+1/50>10/50=1/5
1/51>1/60
1/52>1/60
...
1/60=1/60
=>1/51+1/52+...+1/60>10/60=1/6
=>S>1/4+1/5+1/6=3/5
1/31<1/30
1/32<1/30
...
1/40<1/30
=>1/31+1/32+...+1/40<1/30*10=1/3
1/41<1/40
1/42<1/40
...
1/50<1/40
=>1/41+1/42+...+1/50<10/40=1/4
1/51<1/50
1/52<1/50
...
1/60<1/50
=>1/51+1/52+...+1/60<10/50=1/5
=>S<1/3+1/4+1/5=4/5
\(\dfrac{1}{31}>\dfrac{1}{40}\)
\(\dfrac{1}{32}>\dfrac{1}{40}\)
...
\(\dfrac{1}{40}=\dfrac{1}{40}\)
=>\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}>\dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{10}{40}=\dfrac{1}{4}\)
\(\dfrac{1}{41}>\dfrac{1}{50}\)
\(\dfrac{1}{42}>\dfrac{1}{50}\)
...
\(\dfrac{1}{50}=\dfrac{1}{50}\)
=>\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}>\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{10}{50}=\dfrac{1}{5}\)
\(\dfrac{1}{51}>\dfrac{1}{60}\)
\(\dfrac{1}{52}>\dfrac{1}{60}\)
...
\(\dfrac{1}{60}=\dfrac{1}{60}\)
=>\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{10}{60}=\dfrac{1}{6}\)
=>\(S>\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{3}{5}\)
\(\dfrac{1}{31}< \dfrac{1}{30}\)
\(\dfrac{1}{32}< \dfrac{1}{30}\)
...
\(\dfrac{1}{40}< \dfrac{1}{30}\)
=>\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}< \dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}=\dfrac{10}{30}=\dfrac{1}{3}\)
\(\dfrac{1}{41}< \dfrac{1}{40}\)
\(\dfrac{1}{42}< \dfrac{1}{40}\)
...
\(\dfrac{1}{50}< \dfrac{1}{40}\)
=>\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}< \dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{10}{40}=\dfrac{1}{4}\)
\(\dfrac{1}{51}< \dfrac{1}{50}\)
\(\dfrac{1}{52}< \dfrac{1}{50}\)
...
\(\dfrac{1}{60}< \dfrac{1}{50}\)
=>\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{10}{50}=\dfrac{1}{5}\)
=>\(S< \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{4}{5}\)
=>\(\dfrac{3}{5}< S< \dfrac{4}{5}\)
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
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 \(\dfrac{2}{5}< S< \dfrac{8}{9}\)
S = \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{5}\) + ... + \(\dfrac{1}{8}\) + \(\dfrac{1}{9}\)
Vì \(\dfrac{1}{3}>\dfrac{1}{4}>\dfrac{1}{5}>..>\dfrac{1}{9}\) ta có:
\(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) > \(\dfrac{2}{4}\) = \(\dfrac{1}{2}\)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}>\dfrac{1}{9}.5\) = \(\dfrac{5}{9}>\dfrac{5}{10}=\dfrac{1}{2}\)
Cộng vế với vế ta có:
S > \(\dfrac{1}{2}+\dfrac{1}{2}=1\) (1)
\(\dfrac{1}{3}+\dfrac{1}{4}< \dfrac{2}{3}\)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}< \dfrac{1}{5}.5=1\)
Cộng vế với vế ta có:
\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}\) < \(\dfrac{2}{3}\) + 1 < 2 (2)
Kết hợp (1) và (2) ta có:
1 < S < 2 (đpcm)