# Limits regarding sequences

• Aug 26th 2013, 05:34 PM
Kristen111111111111111111
Limits regarding sequences
I have proven that if {an} and {bn} are 2 sequences such that an >= bn for each n>n0 , for some n0 belongs to positive integers with limit of an when n goes to infinity is L and limit of bn when n goes to infinity is M , then L >= M by using a contradiction proof.

My problem is I think that we can say, if an > bn for each n > n0 ,for some n0 belongs to positive integers, then L > M. But I'm having a difficulty in proving that claim. I have tried to prove it in the same way by using the contradiction proof but I get stuck because then I assume L <= M and as in earlier proof when I get (M-L)/2 as epsilon I get an epsilon which is >= 0 but epsilon should be strictly greater than 0.(Speechless)

So, is there another way of proving this? Please can somebody help?
• Aug 26th 2013, 06:01 PM
Plato
Re: Limits regarding sequences
Quote:

Originally Posted by Kristen111111111111111111
I have proven that if {an} and {bn} are 2 sequences such that an >= bn for each n>n0 , for some n0 belongs to positive integers with limit of an when n goes to infinity is L and limit of bn when n goes to infinity is M , then L >= M by using a contradiction proof.

To prove this by contradiction assume that $\displaystyle M>L$. Then you do know that $\displaystyle \epsilon= \frac{M-L}{2}>0$.

So almost all $\displaystyle b_n>a_n$ .
• Aug 27th 2013, 06:31 AM
emakarov
Re: Limits regarding sequences
Quote:

Originally Posted by Kristen111111111111111111
My problem is I think that we can say, if an > bn for each n > n0 ,for some n0 belongs to positive integers, then L > M.

This is false. Consider a_n = 1/n and b_n = 1/n^2 (ignoring n = 1).
• Aug 28th 2013, 05:35 AM
Kristen111111111111111111
Re: Limits regarding sequences
Well I guess what I thought is wrong when considering your example.. Thank you so much for helping :)
• Aug 28th 2013, 05:46 AM
HallsofIvy
Re: Limits regarding sequences
In general, the best you can say for limits is $\displaystyle \le$ or $\displaystyle \ge$. That is, if $\displaystyle a_n< b_n$ (strictly less than) for all $\displaystyle n> n_0$, then, using the indirect proof that Plato suggested, we can show that $\displaystyle \lim a_n\le \lim b_n$ (less than or equal to).