Limit of Sequence Definition

Question

If we reverse the order of events in the definition of limit of sequence $\langle a(n) \rangle$, we get

There exist $N$ such that for all $\epsilon > 0$ for all $n \geq N$, $|a(n)-a|< \epsilon$.

What does this mean ?

The answer given is "This just means $a(n)=a$ for all $n>N$." I am really not getting idea how is it different from definition of limit. Why can't we define limit like this.

Answer 1

The statement$$(\exists N\in\mathbb N)(\forall\varepsilon>0):n\geqslant N\implies\lvert a-a_n\rvert<\varepsilon$$means that there is a natural number $N$ such that, for any number $\varepsilon>0$, the inequality $\lvert a-a_n\rvert<\varepsilon$ holds whenever $n\geqslant N$. But $\lvert a-a_n\rvert\geqslant0$ and the only non-negative number which is smaller than any number greater than $0$ is $0$. So, what this means is that, if $n\geqslant N$, $\lvert a-a_n\rvert=0$. But$$\lvert a-a_n\rvert=0\iff a-a_n=0\iff a=a_n.$$

Answer 2

The statement

There exist $N$ such that for all $\epsilon>0$ for all $n\ge N,$ we have $|a(n)-a|<\epsilon$

means when you get sufficiently far in the sequence (i.e., from some point $n=N$), all terms after that point are a distance less than $\epsilon$ from a fixed number $a,$ no matter how small $\epsilon$ becomes. That means all terms after that point are arbitrarily close to $a.$ Or, in other words, that they are all $a,$ since that's the only number that's arbitrarily close to $a.$

Answer 3

The condition in your post means that the same $N$ is valid for any $\epsilon$ (whereas in the definition of the limit, $N$ is adjusted as a function of $\epsilon$).

This is virtually the same as saying that $N$ must be valid for $\epsilon=0$, i.e. either $a(n)=a$ or $N$ is infinite (which is not acceptable).