Limits regarding sequences

I have proven that if {a_{n}} and {b_{n}} are 2 sequences such that a_{n} >= b_{n} for each n>n_{0} , for some n_{0} 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 a_{n} > b_{n} for each n > n_{0} ,for some n_{0} 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?

Re: Limits regarding sequences

Quote:

Originally Posted by

**Kristen111111111111111111** I have proven that if {a_{n}} and {b_{n}} are 2 sequences such that a_{n} >= b_{n} for each n>n_{0} , for some n_{0} 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 . Then you do know that .

So almost all .

Re: Limits regarding sequences

Quote:

Originally Posted by

**Kristen111111111111111111** My problem is I think that we can say, if a_{n} > b_{n} for each n > n_{0} ,for some n_{0} belongs to positive integers, then L > M.

This is false. Consider a_n = 1/n and b_n = 1/n^2 (ignoring n = 1).

Re: Limits regarding sequences

Well I guess what I thought is wrong when considering your example.. Thank you so much for helping :)

Re: Limits regarding sequences

In general, the best you can say for limits is or . That is, if (**strictly** less than) for all , then, using the indirect proof that Plato suggested, we can show that (less than or equal to).