Definition of a limit question I am having a hard time understanding the difference between the two statements, 
$$\forall \epsilon >0 \  \exists N \ \in \mathbb{N} \ \forall n\ge N \ (|a_n-a| < \epsilon)$$ and
$$\exists N  \in \mathbb{N} \ \forall \epsilon > 0 \ \forall n\ge N \ (|a_n-a| < \epsilon) $$ 
I can see the difference when I am trying to prove a limit (unless the sequence is constant) because I need to let $N$ depend on $\epsilon$. But when I say the first statement out loud, it seems like it explains what a convergent sequence is doing. 
If there is a $N$ s.t. for any positive number you give me, any index greater than that $N$ implies the distance between our limit and the sequence is less than the positive number. What am I missing?
 A: Notice how this $N$ is enough for every $\epsilon$ given in this second statement. Let's study a sequence that satisfy this second statement.
Let $(a_n)_{n \in \mathbb{N}}$ be a sequence that satisfies it. Then, there exists $N \in \mathbb{N}$ such that for every $\epsilon > 0$ given, $m>N \implies |a_m - a| < \epsilon$. For every $n \in \mathbb{N}$, you can take $1/n$ for $\epsilon$, thus giving you that $m>N \implies |a_m - a| < 1/n \  \forall n \in \mathbb{N} \implies |a_m - a| = 0 \implies a_m = a$. Then it means that $a_n$ is constant for $m>N$!
A: The second definition says that $N$ does not depend on $\varepsilon$: the same $N$ continues to work now matter how small $\varepsilon$ gets.  You want $a_n$ to be within $\varepsilon=0.000000000000000000001$ of $a$?  By the second definition, you don't need to make $N$ any bigger than if you want it to be within $\varepsilon=0.001$.  That will work if after some finite number of steps, i.e. for $n\ge$ some $N$, we have $a_n$ always equal to $a$.  Otherwise then $N$ must depend on $\varepsilon$ and $N$ must get bigger as $\varepsilon$ gets smaller.
