Proving the inequality

  • #1

Main Question or Discussion Point

I encountered this problem in one of my math lecture notebooks; Our professor didnt show how it was done, so that leaves me clueless. The problem was to show that the sequence {a_n} defined by
[tex]
a_1 = 1,
a_2 = \int^2_1 \frac{dx}{x},
a_3 = \frac{1}{2},
a_4 = \int^3_2 \frac{dx}{x} ,
...
[/tex]

When generalized gives For any natural number n,

[tex]
a_{2n-1} = \frac{1}{n}
[/tex]
[tex]
a_{2n} = \int^{n+1}_n \frac{dx}{x} = \ln x |^{n+1}_{n} = \ln \frac{n+1}{n}
[/tex]

is decreasing, that is,
[tex]
\frac{1}{n} > \ln \frac{n+1}{n} > \frac{1}{n+1}
[/tex]

I've tried math induction but I'm stuck at the (ii) part of math induction, and i tried comparing their derivatives, but I can't conclude anything from doing so. I've tried to compute for their areas, but that got me nowhere. I've graphed their functions using a graphing program, and I saw that it is true, but I would like know how i can prove this without graphing...

thanx in advance for all help and advice on my problem
 

Answers and Replies

  • #2
lurflurf
Homework Helper
2,423
123
when a<b
[tex](b-a)\min(f)\leq\int_a^b f dx\leq(b-a)\max(f)[/tex]
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,770
911
Lovely! Absolutely lovely!
 
  • #4
:surprised

whoa... that is absolutely lovely... boy do I need a refresh on my calculus... :rofl:

thanx lurflurf! you're a lifesaver :)
 

Related Threads for: Proving the inequality

  • Last Post
Replies
2
Views
1K
Replies
3
Views
2K
Replies
2
Views
4K
Replies
5
Views
2K
Replies
8
Views
988
Replies
6
Views
3K
Replies
3
Views
1K
Replies
12
Views
2K
Top