TITLE: Convergence proofs: why is it necessary to prove specifically for $\epsilon$?
QUESTION [5 upvotes]: In most proofs of sequence convergence in real analysis (like here and here) in order to prove that, for example 
$$
a_n + b_n \to_n a+b
$$
or 
$$
a_n b_n \to_n ab
$$
when $a_n \to_n a $ and $b_n \to_n b$ often it is often taken that $|a_n -a| < \frac{\epsilon}{2}$, which follows from the fact that (for a different $n$ of course) $|a_n -a|<\epsilon$. Or, for the product proof that $|a_n-a|<\frac{\epsilon}{2(1+a)}$. Once again, for a different $n$. After some algebra it is shown that the desirable quantity, say $a_n b_n$ is within an $\epsilon$ from $a+b$. 
Well here's my confusion (and question): why not just prove that (in the first case)
$$
|a_n +b_n -(a+b)| <2 \epsilon
$$
and this would imply that of course for some other $n$ it is less than $\epsilon$. And the same with other proofs. In short: why, if we are allowed to play around with discrepancies from the limit of other sequences, we absolutely must show that the desired one is strictly less than $\epsilon$ away?

REPLY [2 votes]: It is similar to the difference between leaving a solution un-simplified.  It is clear that answers of $5/10$ and answers of $1/2$ are equivalent but we generally write solutions as simplified as possible.
When it comes to these proofs, an answer showing $|a_n +b_n -(a+b)| < \frac{23486}{777}\epsilon^{239\pi}$ could be considered un-simplified. It is usually a small matter to resolve and should be done if possible. 
If you find that $|a_n +b_n -(a+b)| < 2\epsilon_0^{2}$ you can say "take $\epsilon = 2\epsilon_0^2$" or "take $\epsilon_0 = \sqrt{\frac{\epsilon}{2}}$" which then shows that $|a_n +b_n -(a+b)| < \epsilon.$ It should be clear that resolution of this matter is almost trivial.
Authors of books and papers will write their proofs so that this is unnecessary simply because it produces the cleanest proofs and that is always an advantage when writing anything. Any student should have the skills to take their proofs and modify them to fit this criteria.