CMR: S=1/(2017+1)+1/(2017+2)+...+1/(3.2017+1)>1
Giúp e vs mn ơi 
CMR: S=1/(2017+1)+1/(2017+2)...+1/(3.2017+1)>1
Started By ngocloan, 06-11-2016 - 19:01
-
This topic is locked
#1
Posted 06-11-2016 - 19:01
ngocloan
-
- Thành viên mới
-
- 39 posts
Binh nhất
#2
Posted 06-11-2016 - 20:45
tenlamgi
-
- Thành viên mới
-
- 45 posts
Binh nhất
CMR: S=1/(2017+1)+1/(2017+2)+...+1/(3.2017+1)>1
Giúp e vs mn ơi
Ta có: $S=\sum_{n=2017+1}^{3.2017+1}1/n> \int_{2017+1}^{3.2017+1}dx/x=ln(\frac{3.2017+1}{2017+1})> ln(e)=1$
#3
Posted 07-11-2016 - 18:10
tenlamgi
-
- Thành viên mới
-
- 45 posts
Binh nhất
mk học lớp 8
Ta có:$S=\sum_{n=2017+1}^{2.2017}1/n+\sum_{x=2.2017+1}^{3.2017+1}1/x> \frac{2017^2}{\sum_{n=2017+1}^{2.2017}n}+\frac{2018^2}{\sum_{x=2.2017+1}^{3.2017+1}x}$(BDT Cauchy-Schwarz)
$=\frac{2.2017^2}{(3.2017+1).2017}+\frac{2.2018^2}{(5.2017+2).2018}=\frac{2.2017}{3.2017+1}+\frac{2.2018}{5.2017+2}$
$=4036/10087+2017/3026>1$
- ngocloan likes this
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users
