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A= \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{100}=\frac{99}{100}\)
=> A= \(\frac{99}{100}>\frac{25}{26}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : A = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
\(\Rightarrow A< \frac{8}{9}\)(1)
Lại có : \(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
\(\Rightarrow A>\frac{2}{5}\)(2)
Từ (1);(2) => \(\frac{8}{9}>A>\frac{2}{5}\)
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\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2009.2009}\)
\(\dfrac{1}{2.2}< \dfrac{1}{1.2}=1-\dfrac{1}{2}\)
\(\dfrac{1}{3.3}< \dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\dfrac{1}{4.4}< \dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4}\)
...
\(\dfrac{1}{2009.2009}< \dfrac{1}{2008.2009}=\dfrac{1}{2008}-\dfrac{1}{2009}\)
\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2009.2009}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...\dfrac{1}{2008}-\dfrac{1}{2009}=1-\dfrac{1}{2009}< 1\)
\(\Rightarrow A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{2009.2009}< 1\)
Ta có:
\(\dfrac{1}{2\times2}+\dfrac{1}{3\times3}+\dfrac{1}{4\times4}+...+\dfrac{1}{2009\times2009}< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{2008\times2009}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}=1-\dfrac{1}{2009}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta thấy:
1/2*2<1/1*2)vì 2*2>1*2).
1/3*3<1/2*3(vì 3*3>2*3).
...
1/8*8<1/7*8(vì 8*8>7*8).
=>1/2*2+1/3*3+1/4*4+...+1/8*8<1/1*2+1/2*3+1/3*4+...+1/7*8.
=>B<1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8.
=>B<1-1/8.
=>B<7/8.
Mà 7/8<1.
=>B<1.
Vậy B<1(đpcm).
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 1/[n x (n - 1)] = [(n - 1) - n] / [n x (n - 1)] = 1/n - 1/(n - 1)
Áp dụng : 1/(1x2) + 1/(2x3) + 1/(3x4) + ... + 1/(48x49) + 1/(49x50)
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/48 - 1/49 + 1/49 - 1/50
= 1 - 1/50 < 1
Vậy : 1/(1x2) + 1/(2x3) + 1/(3x4) + ... + 1/(48x49) + 1/(49x50) < 1
Ta có : 1/(n x n) < 1/[(n - 1) x n]
1/(2x2) < 1/(1x2)
1/(3x3) < 1/(2x3)
1/(4x4) < 1/(3x4)
.............
1/(49x49) < 1/(49x49)
1/(50x50) < 1/(49x50)
=> 1/(2x2) + 1/(3x3) + 1/(4x4) + ... 1/(49x49) + 1/(50x50) < 1/(1x2) + 1/(2x3) + 1/(3x4) + ... + 1/(48x49) + 1/(49x50) < 1
Vậy 1/(2x2) + 1/(3x3) + 1/(4x4) + ... 1/(49x49) + 1/(50x50) < 1
Đặt B=1/1*2+1/2*3+...+1/99*100
Ta thấy:
A=1/2*2+1/3*3+...+1/100*100<B=1/1*2+1/2*3+...+1/99*100 (1)
Ta lại có:
B=1/1*2+1/2*3+...+1/99*100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100<1 (2)
Từ (1) và (2) ta có: A<B<1 <=>A<1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+...+\frac{1}{20.20}\)
Ta có: \(\frac{1}{2}-\frac{1}{3}>\frac{1}{3.3}>\frac{1}{3}-\frac{1}{4}\)
\(\frac{1}{3}-\frac{1}{4}>\frac{1}{4.4}>\frac{1}{4}-\frac{1}{5}\)
\(\frac{1}{4}-\frac{1}{5}>\frac{1}{5.5}>\frac{1}{5}-\frac{1}{6}\)
...................................
\(\frac{1}{19}-\frac{1}{20}>\frac{1}{20.20}>\frac{1}{20}-\frac{1}{21}\)
Cộng theo vế ta được:
\(\frac{1}{2}-\frac{1}{20}>S>\frac{1}{3}-\frac{1}{21}\)\(\Rightarrow\)\(\frac{1}{2}>S>\frac{1}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(n.n!=\left(n+1\right).n!-1.n!=\left(n+1\right)!-n!\)
Suy ra \(A=1+1.1!+2.2!+...+10000.10000!\)
\(=1+2!-1!+3!-2!+...+10001!-10000!\)
\(=10001!\)
\(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2021.2021}\)
\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2021^2}\)
Xét : \(\frac{1}{k^2}\left(k\inℕ^∗\right)\)
\(=\frac{4}{4k^2}< \frac{4}{4k^2-1}=\frac{4}{\left(2k-1\right)\left(2k+1\right)}==2\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)\)
Áp dụng cho biểu thức A,ta có :
\(A< 2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{4041}-\frac{1}{4023}\right)\)
\(=2\left(\frac{1}{3}-\frac{1}{4023}\right)=\frac{2}{3}-\frac{2}{4023}< \frac{2}{3}< \frac{3}{4}\)
Yuriko
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