The definition of a sequence converging. The sequence $\{a_n\}$ converges to t he number L if for every positive number $\epsilon$ there corresponds an integer $N$ such that for all $n$,
$$n > N \implies |a_n - L| < \epsilon$$
If no such number $L$ exists, we say that $\{a_n\}$ diverges. If $\{a_n\}$ converges to L, we write $$\lim_{n\to\infty}a_n=L$$
I am having trouble understanding this definition in my textbook. Specifically when they use $\epsilon$, $N$, and $n$. Is $\epsilon$ the values in the sequence, $n$ the n-th term of the sequence, and $N$ an arbitrary integer? Could someone explain this to me? 
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Show that $lim_{n\to\infty}\frac{1}{n}=0$.
The implication will hold if $\dfrac{1}{n}< \epsilon$ or $n > \dfrac{1}{\epsilon}$. If $N$ is any integer greater than $\dfrac{1}{\epsilon}$, the implication will hold for all $n>N$.
 A: Think of it as a game between us:
I choose a small number $\epsilon$ then you have to see if you can find an $N$ such that for all $n > N$ the terms of the sequence are within $\epsilon$ of $L$.
If you can always do so, then the sequence converges - you win.
A: Is this over the real numbers? Let us look at a simple case. Suppose $L=0$. Give me an $\epsilon>0$. What the definition tells me is that I can find a positive integer $N$ such that after the Nth term , that is $n>N$ , $-\epsilon<a_n<\epsilon$. Pictorially, this means that the terms of the sequence eventually get trapped in between the lines $y=\epsilon$ and $y=-\epsilon$. More generally your lines are $y=\epsilon+L$ and $y=L-\epsilon$.
A: The intuitive definition to me is as follows: 
Your sequence $a_n$ converges if: 
"No matter how close to $L$ I want your sequence $a_n$ to be you can find a large index $N$ so that after that point the rest of your sequence is that close $L$"
No matter how close is characterized by the fact that $\epsilon$ can be arbitrarily small. 
