I once saw a proof for convergence for a specific problem involving fixed point iteration showing $|a_n-L| \leq (1/2)^n \rightarrow 0 $ and so $a_n\rightarrow L$
I initially did not understand why they were deciding to abandon $\epsilon-N$ notation now and that it was some sort of handwavy, sketchy explanation since this was in an Applied Mathematics book and not a rigorous Analysis book but realised that it was perfectly legitimate since $|a_n - L| $ is equivalent to $||a_n - L|-0|$- add in the $\epsilon-N$ and that becomes an if and only if.
So in words that's if $a_n$ tends to $L$ then the difference between them gets closer and closer to $0$.
Similarly for showing a sequence is Cauchy: we want all $m,n$ past a certain point to be close enough to each other. So like before I plan to show that $|a_m-a_n| \rightarrow 0$ but my problem here I am dealing with two points and there's infinitely many of them.
Does the following work?
WLOG let $m>n$ ($m=n$ being trivial).
Then $m=n+k$ for some $k>0$
we now want $|a_{n+k}-a_n| < \epsilon$ (assume the stuff about n>N is all there- just boring to keep writing out)
But by the reasoning above this is the same as $|a_{n+k}-a_n| \rightarrow 0$
and by the triangle inequality we know that $|a_{n+k}-a_n|\leq|a_{n+k}-a_{n+k-1}|+|a_{n+k-1}-a_{n+k-2}|+...+|a_{n+1}-a_{n}|$
Now in my specific problem I have managed to find a bound $\forall n$
That is:
$\forall n$ $|a_{n+1}-a_n|\leq c(1/2)^n$ where $c$ is a constant from the problem. So by Alegbra of Limits the whole sum on RHS tends to $0 $and so the left hand side tends to $0$ by sandwiching and so we're done $a_n$ is a Cauchy Sequence.
If this is flawed please could you point out where.
The premise of the question is flawed. The original observation, bounding the error by $(1/2)^n$, is a demonstration of convergence that doesn't need any more $\epsilon$ work to "justify" further. The point of the epsilon notation is to give a rigorous language to discuss asymptotics but here you already have the asymptotic bound in a rather specific form, so no epsilon assistance is required.
With that said, given a strong bound like $(1/2)^n$ it is possible to force it into whatever form a class or textbook on real analysis canonizes as the Only Allowed Protocol for a convergence argument. But if it isn't already fully clear from the bound that convergence has occurred, the class has failed in one of its basic purposes.