Proving a sequence converges $\epsilon-N$ It seems my biggest nightmare has come to haunt me again! I first came across the formal proof of a limit $\epsilon-\delta$ in Calculus 1, I never truly mastered it since at the time it was just racking my brain. I have now begun self-teaching a course in sequences & series and it has already come up. It has also been emphasised to me that being able to understand and do this is fundamental for success in this course. 
I'm almost 100% certain that I'd have no trouble applying the methods, I've seen in tutorials to prove a sequence converges. However, I don't understand how the $\epsilon-N$ proof actually proves a sequence converges; if that makes much sense. 
Here are a few questions, I'd like to ask: 


*

*What does this statement mean: $|a_n - L| < \epsilon $ L refers to the limit. 

*Why does this proof actually prove a sequence converges? 

*Why does $\epsilon>0$?
Those are the only questions that currently come to mind, also if anyone can provide any additional advice on how I can wrap my head around these proofs please feel free to post! 
 A: The definition is the following $$\forall \epsilon >0, \exists n_0\in\mathbb{N} : |a_n-L|\leq \epsilon \,\,\,\forall n\in\mathbb{N}, n>n_0$$
$|a_n-L|$ represents the distance between the value of the limit to which the sequence converges and the $n^{nt}$ element of the sequence. It is a distance in absolute value because $a_n$ could be bigger or smaller than $L$ (depending if the sequence is decreasing or increasing or alternating), so we take the Absolute value to always have a positive value.
$|a_n -L|<\epsilon$ means that this distance is smaller than any value of $\epsilon >0$ no matter how small you take it to be! Recall that if you are doing analysis there should be a lemma saying that if $|x|<\epsilon$, $\forall \epsilon > 0$ $\implies |x| = 0$. Therefore $|a_n -L|<\epsilon$ means that you want this distance to become zero!
The rest of the definition it is saying that if a sequence converges to $L$, i.e. if the $n^{nt}$ term, as $n\to\infty$ goes to $L$ , then this means that after a certain point ($n>n_0$, so $n_0$ is the "turning point") all the terms of the sequence coming after $a_{n_0}$ can approach $L$ so much that the distance between $L$ and the term you've chose, after $a_{n_0}$ is smaller than the given epsilon.
What you do in the proving it, is exactly this: you choose an arbitrary $\epsilon >0$. Then what happens is that you can always find a $n_0$ (which is the subscript of an element of the sequence, say $n_0 = 18923842$, this means that $a_{18923842}$ is the element you're talking about) such that every term after it (i.e. $a_{18923843}$ for instance) has a distance to $L$ which is a value (positive because of the absolute value) smaller than the epsilon you chose before!
