Given $ \lim\limits_{n\rightarrow\infty} na_n= a $, where $a>0$, prove that $\sum_{n=1}^\infty a_n $ diverges

Question

The proof that I was given in class is as follows:

By the definition of the limit: $ \forall \epsilon>0\ $there $\ \exists\ N : |na_n-a|<\epsilon$, when $n>N$.

We have $ a-\epsilon<na_n<a+\epsilon $, when $n>N$

Choose $\epsilon=\frac a2$

Now $\frac {a}{2n}<a_n$

We have that $\sum_{n=1}^{\infty}\frac {a}{2n}$ diverges

And by the comparison test it follows that $\sum_{n=1}^{\infty}a_n$ must diverge.

I have a couple of questions about this proof.

1. Why are we allowed to choose any epsilon? Shouldn't the proof hold for all $\epsilon$? I could have chosen $\epsilon=a$ and comparison test wouldn't hold anymore.

2. Doesn't this prove only the case when $n>N$. My guess it that the sum up to N doesn't influence the divergence/convergence, that's why it holds, but I'd like some verification.

3. More general question. I have seen other proofs where we choose some specific $\epsilon$ in terms of the limits (like $\epsilon = \frac {a+b}{2}$). When are we allowed to do that?