When we do the epsilon delta proof for even the simplest of functions, it is a crucial step to restrict $\delta$ in some set of values before we give the actual value of delta example.
I want to make an analogous result for limit of discrete sequences. Let me show a simple example where we could use that. Suppose we want to prove that $\{\frac{1}{2n+3}\}_n$ converges to $0$. We can prove this in the following way: $\frac{1}{2n+3} <_{(1)} \frac{1}{2n+n}.$ This simplification is justified if $n>3$.
If we are provided that $\{\frac{1}{n}\} $ converges to $0$, then there exist some $\tilde{n}$ such that for $n >\tilde{n}$ , we have $\frac{1}{n} < 3 \epsilon$. Now, clearly any number in the set $\{n | n> \tilde{n} \}$ would satisfy this too. If we have that the intersection $\{n | n> \tilde{n} \} \cap \{ n | n>3\}$ (1) is non empty, we take some $n'$ out of it's intersection and we can chain the inequalities to get: $ \frac{1}{2n+3} < \frac{1}{3n } <\epsilon$.
And this $n'$ we claim then is the required for the $\epsilon-N$ definition for $\frac{1}{2n+3}$.
This trick I mentioned above is used quite often in many problems, but I have never seen justification as to why the intersection (1) is not empty. I am presuming it comes from the fact that both the sets are unbounded from above. Now, I don't know how to prove that either. In particular, I think it'd be a bit difficult to prove from the information I got from this post . Maybe it'd be easy to prove unboundedness of rays in $\mathbb{N}$ when we have that $\mathbb{N}$ itself is unbounded. Or it could be that I am on completely the wrong track.
How do I prove rigorously that the intersection (1) is non empty?