Do sequences with the same tail have equivalence relation and converge to the same limit? I have a problem with the following exercise:
There are two sequences,
$ a_i = -11, -12, -13, -14, -15, -16, -17, -18, -19....$
$ b_i = 1000, 10001, -16, -17, -18, -19...$
And the sequence $ b_i $ from 3rd term would then become the same as $ a_i $ from 6th term, for the rest of the sequences. So in other words, they would have the same tail from 6th term of sequence a and 3rd term of sequence b.
Is that equivalence relation?
I was thinking $ a_{i-3} = b_i $ and $ b_{i-3} = a_i$ for $ i>3$ so it is symmetric and converges to the same limit. The terms are the same so it is reflexive if I understand it correctly. Then it is not transitive. But not sure if that is correct. Also, I believe if one of them converges, then they both converge to the same limit (given sets properties).
Also, what would be the correct definition of such sequences and mathematical notations for them? I am doing Real Analysis as a part of electives for my comp sci degree and am really struggling with the semantics and mathematical language for those..
Appreciate ANY help and suggestions!!
 A: If we wanted to formally define this relation, we might say that two sequences $a_m$ and $b_n$ are "eventually equal" (or maybe "have equal tails") and write, for example, $a_m \sim b_n$ if there exist $M, N \in \mathbb{N}$ such that for all $k \in \mathbb{N}$, $a_{M+k} = b_{N+k}$. In other words, if you remove the first $M-1$ entries from $a_m$ and the first $N-1$ entries from $b_n$ then what's left is the same on both sides.
You can indeed prove that (a) being eventually equal is an equivalence relationship, and (b) sequences that are eventually equal have the same limit. I will sketch out the proofs of both, but I'd encourage you to fill out the details yourself.
First, to prove that $\sim$ is an equivalence relationship, we need to show that it is reflexive, symmetric and transitive.
Symmetric: If $a_m \sim b_n$, then we know that $a_{M+k} = b_{N+k}$ as defined above, so $b_{N+k} = a_{M+k}$ and hence by swapping everything around we should have that $b_n \sim a_m$.
Reflexive: Clearly by choosing $M = N = 0$, $a_{M+k} = a_{N+k}$ so $a_m \sim a_m$.
Transitive: If $a_m \sim b_n$ and $b_n \sim c_l$, then choose $N$ large enough that $b_n$ matches both of the other sequences appropriately and you should be able to show $a_m \sim c_l$.
To prove that equivalent sequences have equal limits, you can use the definition of the limit of a sequence:

*

*$\lim_{n \rightarrow \infty} a_n = L$ if, for any $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that $n > N \implies |a_n - L| < \epsilon$.

If $a_n \sim b_m$ and $\lim_{n \rightarrow \infty} a_n = L$, then for any $\epsilon$ if $|a_n - L| < \epsilon$ beyond some value $N$, then there should be a matching $M$ such that $b_m = a_n$ and hence $|b_m - L| < \epsilon$, thus $\lim_{m \rightarrow \infty} b_m = L$.
