# Proof that $a_n < b_n \implies \lim(a_n) \le \lim(b_n)$

First of all, I'm aware that there are many questions like this on the site, but they all seem to be related to either $$\limsup$$ or $$\liminf$$ and I couldn't find anything that would help me with my problem. I've done some Googling and found some great resources, but I'm still not quite sure how to get to some steps and would like your assistance.

The problem is as follows:

Given $$a_n < b_n$$ prove that $$\lim_{n\to \infty}(a_n) \le \lim_{n\to\infty}(b_n)$$. The proof is then done by contradiction, assuming that $$a = \lim_{n\to \infty}(a_n) > b =\lim_{n\to\infty}(b_n)$$.

We take an $$\epsilon = \frac{a-b} 2$$, so that the $$\epsilon$$-neighborhoods of $$a$$ and $$b$$ are disjoint. From the definition of limits, we now know that there is such a $$N$$, so that $$\forall n > N : |a_n-a|<\frac\epsilon2$$ and $$|b_n-b|<\frac\epsilon2$$.

The next step is absolutely always confusing. Two variants I've found are either:

$$a_n>a-\epsilon=a-\left(\frac{a-b} 2\right)=b+\left(\frac{a-b} 2\right)=b+\epsilon>b_n$$

In which I do not understand why any two terms of that (in)equality are like that, or it is said that if $$a > b$$, there must be such an $$\epsilon$$ so that $$a - \epsilon > b + \epsilon$$. Then, $$a - \epsilon > b + \epsilon > b_n, a_n > a - \epsilon > b + \epsilon > b_n$$, which contradicts $$a_n \le b_n$$. Here I simply cannot comprehend how we came to the conclusion that $$a_n > a - \epsilon$$ and $$b + \epsilon > b_n$$. The definition of the limit uses absolute values everywhere, so surely the values depends on the signs of $$a_n$$ and $$a$$, and $$b_n$$ and $$b$$. Please help me understand what is it that I'm missing here. Thanks in advance!

• $|a|<b \iff -b<a<b$ – user223391 Mar 14 '17 at 18:59

$\newcommand{\eps}{\varepsilon}$If $\eps > 0$ is real, then for every real number $x$, $$\text{|x| < \eps if and only if -\eps < x < \eps.}$$ Particularly, \begin{align*} |a_{n} - a| < \eps &\quad \text{if and only if}\quad -\eps < a_{n} - a < \eps, \\ &\quad \text{if and only if}\quad a - \eps < a_{n} < a + \eps. \end{align*}

Second, if $a$ and $b$ are real numbers such that $b < a$, then $b < \frac{1}{2}(b + a) < a$ (the mean/midpoint lies between). A bit of algebra shows that if $\eps = \frac{1}{2}(a - b)$, then $\eps > 0$, and $$b + \eps = \tfrac{1}{2}(b + a) = a - \eps.$$

In each case, it's a pleasant and instructive exercise to sketch a number line and see the obvious geometric fact these inequalities express.

• I'd like to thank both you and Zachary for the useful insights, I've got it now! And yes, I completely agree that it's both intuitive and obvious when sketched, but I really struggled with finding a formal proof of it. But I guess that's what school is for. Once again, thank you very much for your help! – anonra Mar 14 '17 at 19:21

By definition, ϵ is an arbitrary number that is greater than zero. By expansion of |an−a|<ϵ you get an>a-ϵ and by expansion of |bn-b|<ϵ you get bnb you just have to take ϵ as (a-b)/2 {by using a little bit of your mind} which will be positive as we have assumed a>b. From our previous expansion we have that an>a-ϵ {put in ϵ=(a-b)/2} an>a-ϵ=a- (a-b)/2 which by simple calculation you can show that a - (a-b)/2 = b + (a-b)/2 which is equal to b+ϵ which is greater than bn by our previous expansion. Which overall results that an>bn wheich is a contradiction to our assumption.

First things first: We need to make sure that the limit exists for $a_n$ and $b_n$. by the look of things the book did that by saying letting $a$ and $b$ be those terms.

Now to deal with your issue: Like you said, if the limits do exist then by definition there exist and $N_a\in \mathbb N$ such that $|a_n-a|<\frac {\varepsilon}{2}$ for all $n > N_a$. Likewise this is true for $b_n$. Just keep in mind that the $N$'s might not be the same so I'm going to denote $N_a$ and $N_b$ for the $N$'s of the two sequences. Now we can define $N := max\{N_a,N_b\}$.

This may seem a bit anal, but it is important to understand the distinction now, and not become confused later.

Now, keep in mind that by definition we're able to bound the sequence within a region $\varepsilon$ around $a$. This is what it means when we say $|a_n-a|<\varepsilon$. Any $a_n$ that you can find where $n > N$ will live in the region between $a+\varepsilon$ and $a-\varepsilon$. So specifically $a_n>a-\varepsilon$.

Likewise $b_n<\varepsilon+b$. I do hope this clarifies the issue.

$a_n=0$ for all $n$ and $b_n=\frac{1}{n}$