Understanding a definition of convergence

In addition to the current answer/comment, you are confusing yourself with conflicting notation.

Given the sequence $a_{n} = 1/n$, you then write $$ a_{3}(n) \geq a_{2}(N) \hspace{20pt}\text{and}\hspace{20pt} a_{2}(n) \geq a_{100}(N) $$ but these do not make much sense by your definition. Instead, you would have $$ a_{2} = 1/2, \hspace{20pt} a_{3} = 1/3, \hspace{20pt} \ldots, \hspace{20pt} a_{100} = 1/100 $$ and so on, so that $$ a_{n} = 1/n \hspace{20pt}\text{and}\hspace{20pt} a_{N} = 1/N. $$ Things like $a_{2}(n)$ have not been defined and do not make sense at the moment, which is partially leading to your confusion.

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To flesh out an example, consider the sequence $a_{n} = 1/n$ with $a = 0$. Let $\varepsilon = 1/10$. Clearly if you pick $N = 10$, then $a_{10} = 1/10 \leq \varepsilon$, so $a_{10}$ is in the $\varepsilon$-ball (around $a = 0$). Similarly, we have $a_{11} = 1/11 \leq 1/10 = \varepsilon$, so that $a_{11}$ is also inside the $\varepsilon$-ball. In fact, every $a_{n}$ for $n \geq N = 10$ lies inside the $\varepsilon$-ball.

What if $\varepsilon = 1/100$? Then we must pick $N = 100$ so that every $a_{n}$ for $n \geq N$ lies inside the $\varepsilon$-ball. It's usually a bit of work to determine how $N$ depends on $\varepsilon$ (in this case it is simply $N = 1/\varepsilon$), but once you can show that $N$ depends on $\varepsilon$ in some way so that that definition holds, you've shown convergence.

$(a_{n}) \rightarrow a$ if for every $\epsilon>0 , \exists N \in \mathbb{N}$ such that whenever $n \geq N$ it follows that $|a_{n}-a|< \epsilon$

• first you pick a (positive) margin of error, $\epsilon>0$;
• then you find an index $N$ such that $|a_{n}-a|< \epsilon$ is satisfied at least for all indices $n \ge N$.

Now if you can find such an $N$, which may depend on $\epsilon$, for any positive $\epsilon$ you pick, then we say that $(a_n)$ converges to $a$, i.e. $(a_{n}) \rightarrow a$.

does it mean that this $N$ is the index ?

So $N$ plays the role of a "boundary index": the inequality $|a_{n}-a|< \epsilon$ has to be satisfied for all indices beyond this $N$. If you pick a smaller $\epsilon$, you may need to increase $N$ accordingly.

does it mean that this $N$ is the index ? so we have that $a_{3}(n) \geq a_{2}(N)$

This notation is a bit weird: $a_n$ already means the $n$th element in the sequence, what is $a_3(n)$...?

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 $0$.