S= 1+1/2^2+1/3^2+...+1/2005^2
Chứng minh S<1
K
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B
0
M
3
17 tháng 5 2021
`M=1/2^2+1/3^2+1/4^2+...+1/2021^2`
Vì `1/2^2>1/(2.3)`
`1/(3^2)>1/(3.4)`
`....................`
`1/2021^2>1/(2021.2022)`
`=>M>1/(2.3)+1/(3.4)+............+1/(2021.2022)`
`=>M>1/2-1/3+1/3-1/4+..........+1/2021-1/2022`
`=>M>1/2-1/2022=505/1011=1/3+56/337>1/3(1)`
Vì `1/2^2<1/(1.2)`
`1/(3^2)<1/(2.3)`
`....................`
`1/2021^2<1/(2021.2020)`
`=>M<1/(1.2)+1/(2.3)+............+1/(2020.2021)`
`=>M<1-1/2+1/2-1/3+..........+1/2020-1/2021`
`=>M<1-1/2021<1(2)`
`(1)(2)=>1/3<M<1`
17 tháng 5 2021
+Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3};\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4};\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5};...;\dfrac{1}{2021^2}=\dfrac{1}{2021.2021}>\dfrac{1}{2021.2022}\)\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2021.2022}=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2021}-\dfrac{1}{2022}=\dfrac{1}{2}-\dfrac{1}{2022}=\dfrac{505}{1011}>\dfrac{1}{3}\left(1\right)\)+Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2};\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{2021^2}< \dfrac{1}{2020.2021}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020.2021}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}=1-\dfrac{1}{2021}< 1\left(2\right)\)Từ (1) và (2) suy ra: \(\dfrac{1}{3}< M< 1\)
KW
0
18 tháng 7 2019
Ta có :
S= 1/51 +1/52 +..+1/100
Vì 1/51>1/52>...>1/100
=> S >1/100 * 50 =1/2 (1)
Vì 1/100 <1/99<...<1/51<1/50
=> S < 1/50 * 50=1 (2)
Từ (1),(2) => 1/2 < S<1
P=1/2^2+1/2^3+...+1/2^2018
2P=1/2 +1/2^2 +...+1/2^2017
=> 2P-P= (1/2 +1/2^2 +...+1/2^2017)-(1/2^2+1/2^3+...+1/2^2018 )
=> P=1/2 -1/2^2018 <1/2 <3/4
K
18 tháng 7 2019
Ta có: \(\frac{1}{51}>\frac{1}{100};\frac{1}{52}>\frac{1}{100};...;\frac{1}{100}=\frac{1}{100}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>\frac{1}{100}.50=\frac{1}{2}\)
\(\Rightarrow S>\frac{1}{2}\)
Ta có \(\frac{1}{51}< \frac{1}{50};\frac{1}{52}< \frac{1}{50};...;\frac{1}{100}< \frac{1}{50}\)
\(\Rightarrow\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}< \frac{1}{50}.50=1\)
\(\Rightarrow S< 1\)