In showing the convergence of a sequence, does it matter how one chooses $n$? I am doing an exercise in Abbott's Understanding Analysis textbook (question 2.2.1.b). 

I want to show that: $\displaystyle\lim\frac{3n+1}{2n+5}=\frac{3}{2}.$

Now, i've begun by first working with the inequality I need: $\displaystyle\left|\frac{3n+1}{2n+5}-\frac{3}{2}\right| < \epsilon.$ 
Simplifying the LHS, I get the inequality: $\displaystyle\left|\frac{3n+1}{2n+5}-\frac{3}{2}\right|=\frac{13}{2|2n+5|} < \epsilon.$
Now my reasoning is as follows. If I start with an $\epsilon$ larger than $\displaystyle\frac{13}{2|2n+5|}$, certainly it would satisfy the original inequality I want. Then I can simply solve for $n$ here to find out how big I need $n$ to be given some $\epsilon$.
Now, testing it out I chose $\epsilon=0.01$, my inequality tells me I need $n>325$ so that my sequence is in this $\epsilon$ neighborhood.
But with a simpler inequality, I can have: $$\displaystyle\left|\frac{3n+1}{2n+5}-\frac{3}{2}\right|<\left|\frac{3n+1}{2n}-\frac{3n}{2n}\right|=\frac{1}{2n}<\epsilon.$$
Now I notice that with this more "optimal" inequality, I just need $n>50$. Of course my first choice with $n>325$ would allow my sequence to be in the $\epsilon$ neighborhood I wanted, but $n>50$ for the first one isn't good enough.
Technically, does it matter how one chooses the inequality? would the first one have been proper?

Either one would be fine. As long as you prove that for sufficiently large values of n, $\frac{3n+1}{2n+5}<\epsilon$ then you're done. It doesn't matter how large $n$ needs to be. As long as it is proved for sufficiently large $n$, no matter what the bound is the proof is fine.
However, on re-look there is a mistake in your second proof. You say
$$\displaystyle\left|\frac{3n+1}{2n+5}-\frac{3}{2}\right|<\left|\frac{3n+1}{2n}-\frac{3n}{2n}\right|=\frac{1}{2n}$$
but actually, $\left|\frac{3n+1}{2n+5}\right|<\left|\frac{3}{2}\right|$ for sufficiently large $n$, which means that the LHS is actually $-\left(\displaystyle\frac{3n+1}{2n+5}-\frac{3}{2}\right)$. I can't see many ways to make the limit much simpler to evaluate (unless you divide the numerater and denominator by $n$).
