1/5+1/13+1/25+...+1/n^2+(n+1)^2 <1/2
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BN
26 tháng 2 2018
Chứng minh 1 bất đẳng thức cơ bản sau:\(\dfrac{1}{n^2+\left(n+1\right)^2}< \dfrac{1}{2n\left(n+1\right)}\)
Thật vậy: \(\dfrac{1}{n^2+\left(n+1\right)^2}=\dfrac{1}{n^2+n^2+2n+1}=\dfrac{1}{2n^2+2n+1}< \dfrac{1}{2n^2+2n}=\dfrac{1}{2n\left(n+1\right)}\)
Thay vào bài toán \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^2}=\dfrac{1}{1^2+\left(1+1\right)^2}+\dfrac{1}{2^2+\left(2+1\right)^2}+\dfrac{1}{3^2+\left(3+1\right)^2}+...+\dfrac{1}{n^2+\left(n+1\right)^2}\)
\(< \dfrac{1}{2.1.2}+\dfrac{1}{2.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2n\left(n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}-\dfrac{1}{2\left(n+1\right)}< \dfrac{1}{2}\left(đpcm\right)\)
TS
0
3 tháng 4 2019
Ta có:\(n^2+\left(n+1\right)^2=n^2+n^2+2n+1=2n^2+2n+1>2n^2+2n=2n\left(n+1\right)\)
\(\Rightarrow\frac{1}{n^2+\left(n+1\right)^2}< \frac{1}{2n\left(n+1\right)}\)
Áp dụng vào bài toán,ta có:
\(\frac{1}{1^2+2^2}+\frac{1}{2^2+3^2}+\frac{1}{3^2+4^2}+......+\frac{1}{n^2+\left(n+1\right)^2}\)
\(< \frac{1}{2\cdot1\cdot2}+\frac{1}{2\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+.....+\frac{1}{2\cdot n\cdot\left(n+1\right)}\)
\(=\frac{1}{2}\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{n\left(n+1\right)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{n}-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}-\frac{1}{2\left(n+1\right)}\)
\(< \frac{1}{2}\)
Y
8 tháng 4 2023
đặt A=\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{41}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{113}\)
= \(\dfrac{1}{5}+(\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{41})+(\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{113})\)
=> A< \(\dfrac{1}{5}+(\dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12})+(\dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60})\)
A<\(\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\)=\(\dfrac{1}{2}\)
vậy A<\(\dfrac{1}{2}\),<2=> A<2 (đpcm)
3 tháng 4 2019
1
B= 12/1.4.7 + 12/4.7.10 + 12/7.10.13 + ... + 12/54.57.60
=> 1/2B= 6/1.4.7 + 6/4.7.10 + 6/7.10.13 + ... + 6/54.57.60
=> 1/2B = 1/1.4 - 1/4.7 +1/4.7 - 1/7.10 +1/7.10 - 1/10.13 + ... + 1/54.57 - 1/57.60
=> 1/2B =1/1.4 - 1/57.60
=> 1/2B = 1/4 - 1/3420
=> 1/2B = 427/1710
=> B = 427/1710 . 2
=> B = 427/855
3 tháng 4 2019
2
A= 1+ 1/22 + 1/32 +...+1/1002
=1+ 1/2.2 + 1/3.3 +...+ 1/100.100
=> A< 1+ 1/1.2 + 1/2.3 +...+ 1/99.100
= 1+ 1 - 1/2 +1/2 - 1/3 +...+1/99 - 1/100
= 2- 1/100 < 2
Vậy A < 2
M
15 tháng 11 2017
Ta có :
\(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
...................
\(\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\).
\(\Leftrightarrow\frac{1}{1^2}+\frac{1}{2^2}+....+\frac{1}{n^2}< \frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{\left(n-1\right).n}\)
\(\Leftrightarrow\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2}< 1+1-\frac{1}{2}+\frac{1}{2}-....+\frac{1}{n-1}-\frac{1}{n}\).
\(\Leftrightarrow\frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2}< 2-\frac{1}{n}\)
\(\Rightarrowđpcm\)
BT
15 tháng 11 2017
Gọi vế trái là A. Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2};\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3};....;\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}=\frac{1}{n-1}-\frac{1}{n}.\)
=> \(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
=> \(A< 2-\frac{1}{n}\) (ĐPCM)