# Thread: Sequences and Limits Proof

1. ## Sequences and Limits Proof

let {xn}n tends to infinity and {yn}n tends to infinity be two convergent (in R) sequences such that xn>or=yn for all natural numbers n. Show that:

as both tend to infinity lim xn>or=lim yn

I'm doing some review and came across this question, I have on which is similar except instead of limits it's for the supremum of the sequences. Would the proofs be similar?

2. Originally Posted by amm345
let {xn}n tends to infinity and {yn}n tends to infinity be two convergent (in R) sequences such that xn>or=yn for all natural numbers n. Show that:

as both tend to infinity lim xn>or=lim yn

I'm doing some review and came across this question, I have on which is similar except instead of limits it's for the supremum of the sequences. Would the proofs be similar?

Very simmilar. Try it and if you get stuck write back and say where.

Tonio

3. Alright, I tried it but I could not seem to get anywhere. Also the proof I had done for the supremums was in extended reals, not just reals, although that just rules out infinity. I'm not sure why but I'm just having a hard time with this. Can you provide some more guidance?

4. Originally Posted by amm345
Alright, I tried it but I could not seem to get anywhere. Also the proof I had done for the supremums was in extended reals, not just reals, although that just rules out infinity. I'm not sure why but I'm just having a hard time with this. Can you provide some more guidance?

This is very basic stuff though annoying to write it and show it , but let's see:

== First, if $\displaystyle \{x_n\}$ is a convergent sequences of non-negative terms then $\displaystyle \lim_{x\to \infty}x_n:=X\geq 0$, otherwise: choose $\displaystyle \epsilon=\frac{-X}{2}$ (note that this number is positive), then $\displaystyle \exists\,N\in\mathbb{N}\,\,s.t.\,\,\forall\,n>N\,, \,|x_n-X|<\epsilon=\frac{-X}{2}\,\,\Longleftrightarrow\,\, -\epsilon=$ $\displaystyle \frac{X}{2}<x_n-X<\frac{-X}{2}=\epsilon\,\,\Longrightarrow \,\,x_n<\frac{-X}{2}+X=\frac{X}{2}<0$, which contradicts the data that $\displaystyle \{x_n\}$ is a non-negative sequence

== Second...we're done, since $\displaystyle \{x_n-y_n\}$ is a non-negative sequence converging (by arithmetic of limits) to $\displaystyle \lim_{n\to\infty}x_n+\lim_{n\to\infty}y_n$...

Tonio

5. Oh I figured it out! I should've posted back and saved you time! Sorry! but thank you.