Hard time understanding what the significance of the tail of a sequence is? I know for the formal definition you have to let $\varepsilon>0$ and there has to exist some $N$ such that $$\exists N \text{ s.t. } n \geq N,$$ and that $$|a_n - L| < \varepsilon$$
What exactly is a tail of a sequence and what does "there has to exist some $N$ such that $n\ge N$" mean? After reading my textbook and many other online resources I have no answer. 
 A: The definition does not say there exists $N$ such that $n\ge N$ and $|a_n-L|<\varepsilon.$
It says there exists $N$ such that for every $n\ge N,$ we have $|a_n-L|<\varepsilon.$
Say you want to make the terms of the sequence differ from $L$ by less than $0.000001.$
That means you're setting $\varepsilon= 0.000001.$
Maybe the first billion terms of the sequence are not that close to $L,$ or not all of them, but after the billionth term, every term of the sequence does come at least that close to $L.$
That means $N=(\text{1 billion} + 1)$ is big enough.
Every term after the billionth term means every term $a_n$ for which $n\ge (\text{1 billion} + 1),$ i.e. $n\ge N.$
The sequence of terms after the billionth term is a "tail" of the sequence.
A: The two displayed lines in the Q should be combined into a single phrase that says:
For any $e>0$ there exists $N\in \Bbb R$  such  that  $\forall n\geq N\;(|a_n-L|<e).$
This the  definition of $\lim_{n\to \infty}a_n=L,$ and the def'n of the phrase "$a_n$ converges to $L$ as $n\to \infty$."
Equivalently, $\lim_{n\to \infty}a_n=L$ iff for any $e>0,$ the set $\{n\in \Bbb N: |a_n-L|\geq e\}$ is finite.
It is not necessary that $N\in \Bbb N.$ 
When the sequence $(a_n)_{n\in \Bbb N}$ converges, the smallest possible $N\in \Bbb N$ usually depends on $e.$ Some students get bogged down in details trying to find the smallest possible $N\in \Bbb N$ (for a given $e$ ) when trying to prove that  a  sequence converges to a value $L$, not noticing that all you need  to prove is that $some $ $ N$ exists that fulfills the conditions.. That is if $n\geq N\implies |a_n-L|<e,$ then for any $N'>N$ we also have $n\geq N'\implies |a_n-L|<e.$
For example to show that $(a_n)_{n\in \Bbb N}$ converges to $1$ when $a_n=1-\frac {n}{2n^2-1}$ for each $n\in \Bbb N,$  observe that $n\in \Bbb N\implies 2n^2-1=n^2 +(n^2-1)\geq n^2+(n-1)\geq n^2>0.$ 
So $0<\frac {n}{2n^2-1}\leq\frac {n}{n^2}=\frac {1}{n}.$ So  if $e>0$ and $N=1+\frac {1}{e}$ then $n\geq N\implies \frac {1}{n}<e\implies |a_n-1|<e. $
Equivalently, in this example, if $e>0$ then the set $\{n\in \Bbb N:|a_n-1|\geq e\}$ is finite because it is a subset of the finite set $\{n\in \Bbb N: n<1+\frac {1}{e}\}.$
A sequence can be defined as a function whose domain is $\Bbb N.$ So if we refer to a sequence $(A_n)_{n\in \Bbb N}$ what we are saying is that $A_n=f(n)$ for some function $f.$ But in many circumstances this can be notationally awkward. I often abbreviate as  $(A_n)_n.$ Some writers and textbooks abbreviate as $(A_n)$ or $\{A_n\}.$ The latter  should not be confused with the usual set-theoretic notation for a set.
