Define an equivalence relation on integer functions $a, b: \Bbb{Z} \to \Bbb{Z}$ by $a \sim b \iff \{a(n) - b(n) : n \in \Bbb{Z}\} = \{0\}$ or is finite (not infnite).
If $a \sim b$ and $b \sim c$, then if $Q = \{a(n) - c(n)\}$ is infinite, then then if $R= \{b(n) - c(n)\}$ is finite, we have that $S = \{a(n) - b(n)\}$ must be infinite since $Q = \{ x(n) + y(n) : x = a - b, y = b - c\} \subset R + S$ is a n elementwise summation of two finite sets otherwise.
Thus $\sim$ is transitive. It is clearly reflexive since if $a - b$ is infinite then so is the set of negative images thereof or $b - a$. In other words it's a symmetric relation.
It is reflexive because $a(n) - a(n) = 0$ and we have that $= \{0\}$ condition in there.
Does the set of equivalence classes $X$ respect addition of sequences in a well-defined manner, i.e. does defining:
$$ [a] + [b] := [a + b] $$
mean that if $[a] \sim [a'], [b] \sim [b']$ that $[a + b] \sim [a' + b']$?
If $(a' + b') - (a + b)$ is not identically $0$, and both $a' - a$ and $b' - b$ have finite image then the image $\{(a' - a)(n) + (b' - b)(n)\}$ is finite being a subset of the elementwise sum of two finite sets $\{(a' - a)(n)\}$ and $\{(b' - b)(n)\}$. $\square$.
I hope that proves the question. So my next question is what is the name of the group of equivalence classes $X$ under such addition?
Suppose that $a\sim a'$ and $b\sim b'$. Let $$D_a=\{a(n)-a'(n):n\in\Bbb Z\}$$ and $$D_b=\{b(n)-b'(n):n\in\Bbb Z\}\;.$$ Then for any $n\in\Bbb Z$ we have
$$(a+b)(n)-(a'+b')(n)=a(n)-a'(n)+b(n)-b'(n)\in D_a+D_b\;,$$
where as usual $D_a+D_b=\{k+\ell:k\in D_a\text{ and }\ell\in D_b\}$. Clearly $D_a+D_b$ is finite; in fact $|D_a+D_b|\le|D_a||D_b|$. Thus, $a+b\sim a'+b'$, and we may define addition of $\sim$-equivalence classes by $$[a]+[b]=[a+b]\;.$$