TITLE: Understanding a definition of convergence
QUESTION [8 upvotes]: I have been trying to understand the definition of convergence, but one small detail in the definition is bother me. The detail is '$n \geq N$'
So the definition I am using is 
$(a_{n}) \rightarrow a$ if for every $\epsilon>0 , \exists N \in \mathbb{N}$ such that whenever $n \geq N$ it follows that $|a_{n}-a|< \epsilon$
So in my view it seems like a sequence converges if you can pick a point in the sequence s.t. after this point in the sequence the sequence will KEEP staying inside the epsilon neighborhood. 
Just to stay concrete suppose $a_{n}=\frac{1}{n}$
then when I look at ''$\exists N \in \mathbb{N}$ s.t. whenever $n \geq N$'' I get confused. 
If I pick $\epsilon = \frac {1}{10}$ Then the sequence is inside the epsilon neighborhood whenever $n>10$ and here my question is what does $n \geq N$ means?
does it mean that this $N$ is the index ? so we have that $a_{3}(n) \geq a_{2}(N)$
or is the meaning that the output you get when you plugin the number into the sequence so we have $a_{2}(n) \geq a_{100}(N)$? because $1/2 \geq 1/100$?

REPLY [2 votes]: To understand the definition and be able to work with it properly it may be helpful to use an equivalent definition that has one variable less.  Namely, a sequence $(a_n)$ converges to $a$ if for every $\epsilon>0$ there is an $N\in \mathbb N$ such that all the terms
$$
a_N, a_{N+1}, a_{N+2}, \ldots
$$
are within $\epsilon$ of $a$.  Thus $\frac{1}{n}$ coverges to $0$ because for each $\epsilon>0$, all the terms starting at rank $\left\lceil\frac{1}{\epsilon}\right\rceil$ will be within $\epsilon$ of $0$.