Converse of order limit theorem

Question

Part of the Order Limit Theorem states that:

Assume $(u_n) \rightarrow l$ and $(v_n) \rightarrow m$, then if $v_n \le u_n$ for all $n \ge N \in \mathbb{N}$, then $m \le l$.

I can prove this. However, I was just wondering whether the "opposite" is also true, that is:

Assume $(u_n) \rightarrow l$ and $(v_n) \rightarrow m$, if $m < l$ then there exists $N \in \mathbb{N}$ such that for all $n \ge N$, $u_n > v_n$.

If true, how do I prove it?

Answer 1

Using the same notation as yours.

Let define $a_n= u_n-v_n$ so by the limit laws we already know that $a_n \to a$ where $a=l-m>0$. So we need to show that there is some $N$ such that $a_n>0$ for all $n\ge N$.

Setting $\varepsilon=a/2$, there is a $N$ such that for all $n\ge N$ we have $|a_n-a|< \varepsilon$, but this means $a_n>a-\varepsilon=a/2>0$. Thus for all $n\ge N$ we may have $a_n>0$, i.e., $u_n-v_n>0$ as was to be shown.

Answer 2

Note that the sequence $(u_n-v_n)\to l-m$ and $0<l-m$. What happens if you take $l-m=\varepsilon$ in the definition of limit of a sequence (for the sequence $(u_n-v_n)$)?

Answer 3

NOTE: I was a bit sloppy in my initial proof; I've added some absolute values.

Suppose not. Let $\epsilon = \frac{l-m}{2}$. Then there exists $N_{1}, N_{2}$ such that $\forall n > N_{1}$, $|l-u_{n}| < \epsilon$. Similarly, $\forall n > N_{2}$, $|v_{n}-m| < \epsilon$. Now, pick $N = max(N_{1}, N_{2})$. Then, $l-m \leq (l-u_{n}) + (v_{n}-m) \leq |l-u_{n}| + |v_{n} - m| < 2\epsilon = l-m$. This is a contradiction. Note the inequality is due to supposing $u_{n}-v_{n} \leq 0$.