# If $a_n$ and $b_n$ are converging sequences, and for all $n\in\mathbb Z+$, $a_n \leq b_n$, then $\lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n$. [duplicate]

I am having a bit of trouble answering this question. If $a_n$ and $b_n$ are converging sequences, and for all $n \in\mathbb Z+$, $a_n$ ≤ $b_n$, then $\lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n$.

Letting $\epsilon>0$, I began with showing that there exists $N_1\in \mathbb N$ such that for all $n \in \mathbb N$, if $n>N_1$, then $|a_n - L|< \epsilon$.

Similarly exists $N_2\in \mathbb N$ such that for all $n \in \mathbb N$, if $n>N_2$, then $|b_n - M|< \epsilon$.

However from here I do not know how to proceed. Do I use the triangle inequality somehow?

## marked as duplicate by Martin R, Nosrati, user91500, José Carlos Santos, Delta-uSep 20 '18 at 11:06

By contradiction, if $L\gt M$, let $\epsilon=\frac{L-M}2\gt0$. Then for $n\gt \operatorname{max}(N_1,N_2)$, we have $a_n\gt L-\epsilon=\frac{L+M}2= M+\epsilon\gt b_n$.

Lat $l_a=\lim_{n\to\infty}a_n$ and $l_b=\lim_{n\to\infty}b_n$. You want to prove that $l_a\leqslant l_b$. Suppose otherwise, that is, suppose that $l_a>l_b$. Now, take $\varepsilon=\frac{l_a-l_b}2$. Then, take $N\in\mathbb N$ such that$$n\geqslant N\implies\lvert a_n-l_a\rvert,\lvert b_n-l_b\rvert<\varepsilon.$$Then, if $n\geqslant N$,$$a_n>l_a-\frac{l_a-l_b}2=\frac{l_a+l_b}2\text{ and }b_n<l_b+\frac{l_a-l_b}2=\frac{l_a+l_b}2,$$and, in particular, $a_n>b_n$, which is a contradiction.

$z_n:= b_n-a_b$ , and

$z:= \lim_{n \rightarrow \infty}b_n -\lim_{n \rightarrow \infty} a_n$.

Need to show that $z \ge 0$.

Assume $z \lt 0$.

For $\epsilon = -z/2 >0$ there is a $n_0 \in \mathbb{Z^+}$

s.t. for $n\ge n_0$: $|z_n -z| \lt \epsilon$

$-\epsilon +z <z_n < \epsilon +z = -z/2+z=$

$z/2 < 0$.