Convergent sequence: the step "for n > N" Below is a proof from the book "How to think about Analysis" by Lara Alcock. 
I'm bad at proofs, but working on it. So in the proof below I miss the reasoning for the line in the red rectangular. Could you please help me with it?
So, we found N in dependence of epsilon, a value that will do for every epsilon. How do we use this to show that for every n > N all terms are within epsilon of 3? 

 A: We know $n>N=\bigl\lceil\frac4\varepsilon\bigr\rceil\ge\frac4\varepsilon$, then$$n>\frac4\varepsilon\implies n\varepsilon>\frac4\varepsilon\varepsilon=4\implies \varepsilon=\frac{n\varepsilon}n>\frac4n$$
I hope the absolute value is clear, but just for completeness:
$$\biggl|3-\frac4n-3\biggr|=\biggl|-\frac4n\biggr|=\biggl|\frac4n\biggr|=\frac4n$$
because $\frac4n$ is positive.
A: if $n>N$, $$n>\frac{4}{\varepsilon}\implies \frac{4}{n}<\varepsilon.$$
Therefore
$$|a_n-3|=\left|3-\frac{4}{n}-3\right|=\left|-\frac{4}{n}\right|=\frac{4}{n}\underset{because\ n>N}{<}\varepsilon.$$
A: When reading the proof, at the end, I felt like the job have beed done twice and that confused me. My problem was not with calculation, but with the reasoning and connection between steps of the proof. I think, now I understand it. 
The aim is to show that:
$\forall \epsilon \gt 0 \ \exists N\in \mathbb N \ such \ that \ \forall n \gt N,\lvert  (3-\frac{4}{n}) - 3\rvert \lt \epsilon $
First, we find N to show that there exists such N. $\lvert-\frac{4}{n}\rvert < \epsilon \Rightarrow \frac{4}{n} < \epsilon \iff n > \frac{4}{\epsilon}, lets\ say \ our \ N = \lceil \frac{4}{\epsilon} \rceil$
Here, at this step, we did the calculation we need, we got our $\frac{4}{n} < \epsilon \iff n > \frac{4}{\epsilon}$. And we found N for which it holds. And our N is already larger than $\frac{4}{\epsilon}$ because we applied a ceiling function on it. And then we wanna prove the last part of the claim:
$\forall n \gt N,\lvert  (3-\frac{4}{n}) - 3\rvert \lt \epsilon$
Of course this is true as we already found out that for this to be true $n > \frac{4}{\epsilon}$ must hold, and we said $N = \lceil \frac{4}{\epsilon} \rceil > \frac{4}{\epsilon}$, so for any value greater than N this is also true. Hence, [here comes the claim].
This was the kind of explanation I missed. If I'm not mistaken, it seems like I got it.
Thank you very much for your time and efforts to help me.
A: To prove that $\displaystyle 3-\frac{4}{n}$ comes arbitrary close $(\varepsilon>0)$ to $3$ when $n$ is big enough, $N$ was selected so that for any $n>N$, $|(\displaystyle 3-\frac{4}{n})-3|$ become less than $\varepsilon$. 
