Is this a valid way to prove that a sequence $\{a_n\}$ given by $a_n=\frac{1}{\sqrt{n+3}} + \frac{n-3n^2}{2n^2-1}$ converges? I'll be using the Archimedean Property to prove this.
Proof
Let $\epsilon>0$. Then $\epsilon^2/4>0$.
By the Archimedean Property, $\forall \epsilon^2/4>0, \exists N\in \mathbb{N}$ such that $1/N<\epsilon^2/4$.
($\star$) Since $1/N<\epsilon^2/4$ it follows that $2/\sqrt{N}<\epsilon$.
$|a_n+\frac{3}{2}|=|\frac{1}{\sqrt{n+3}} + \frac{n-3n^2}{2n^2-1}+\frac{3}{2}|=|\frac{1}{\sqrt{n+3}}+\frac{2n-3}{4n^2-2}|<|\frac{1}{\sqrt{n}}+\frac{n}{2n^2-1}|<\frac{2}{\sqrt{n}}$
$\frac{2}{\sqrt{n}}<\frac{2}{\sqrt{N}}\forall n\geq N$
By ($\star$), $|a_n+\frac{3}{2}|<\epsilon$ $\forall n\geq N$.
Therefore the sequence $\{a_n\}$ converges to $-\frac{3}{2}$.

I don't see why you need that
$\epsilon^2/4$ statement.
Once you have shown that
$|a_n+\frac{3}{2}|<\frac{2}{\sqrt{n}}
$,
to make 
$|a_n+\frac{3}{2}|
\le \epsilon$,
just choose
$n$ so that
$\frac{2}{\sqrt{n}}
\lt \epsilon$
or
$n > 4/\epsilon^2$.
