Having problems with the actual proof step of epsilon-N sequence convergence proofs I'm having problems actually completing epsilon-N proofs for the convergence of sequences, namely the part where, once I've found an N, I have to work backward to derive the original statement. I almost always up ending in a situation where I simply can't reverse one of the steps, logically speaking.
For instance, what I assume is a very easy example, that $\dfrac{2n-1}{4n^2}$ converges to $0$ and thus $\left|\dfrac{2n-1}{4n^2}\right| < \varepsilon $ for all $\varepsilon > 0$.
Finding an $N$ is easy enough:
$\left|\dfrac{2n-1}{4n^2}\right| < \varepsilon $ -- Assuming $n>0$ we can drop the absolute value: 
$\dfrac{2n-1}{4n^2} < \varepsilon $ -- Assuming $n>1$ we can create a smaller fraction. This was a method gone over in class:
$\dfrac{1}{4n^2} < \dfrac{2n-1}{4n^2} < \varepsilon $ -- inverting and simplifying:
$n > \dfrac{1}{2\sqrt{\varepsilon}}$
So then I choose $N = \dfrac{1}{2\sqrt{\varepsilon}}$ and reverse the steps, which is all doable and straightforward up to $\dfrac{1}{4n^2} < \varepsilon $.
But at this point I'm stuck, because creating a smaller fraction only works one way; even if $\dfrac{1}{4n^2} < \varepsilon $, that doesn't mean I can make the left side larger and have it still be true. I don't see any other way to get back to the original statement, though. It's possible/likely I made a mistake deriving some part of this, but it happens with virtually every question I try where I have to use the "create a smaller quantity" method.
Thanks in advance.

$a_n = \frac {2n-1}{4n^2}$
If $n>1$ then $0<a_n< \frac {1}{2n}$
For any epsilon, if $N > \frac {1}{2\epsilon}$ then $n>N \implies |a_n| < \epsilon$ 
