How many sequences of rational numbers converging to 1 are there?

The number of sequence of rational numbers converging to $1$ is not countable. Suppose you get all squences by $(a_n^{(1)})_{n\in \mathbb{N}}, (a_n^{(2)})_{n\in \mathbb{N}}, (a_n^{(3)})_{n\in \mathbb{N}}, \dotsc$ Define a sequence $(b_n)_{n\in \mathbb{N}}$ by $$b_{k}:=\begin{cases}1 & a_k^{(k)}\neq 1\\ 1+\frac{1}{n}& a_k^{(k)}= 1 \end{cases}$$ Then $\lim_{n\to\infty}b_n=1$ but $(b_n)_{n\in \mathbb{N}}\neq (a_n^{(i)})_{n\in \mathbb{N}} $ for all $i\in\mathbb{N}$.

For each real number $x$ define $$a_n(x):= 1+ \frac{ \lfloor xn \rfloor}{n^2}$$

Show that this is a one-to-one function from $\mathbb R$ to the set of sequences converging to $1$.

I believe there is a continuum number of such sequences. For ease of writing, let's try to equivalently count the number of sequences converging to $0$. First note that there are countably many rational numbers in the interval $[-a,a]$. Now set for every sequence we will consider $x_i \in [\frac{-1}{i},\frac{1}{i}]$, so that each sequence of $x_i$'s will converge to $0$. The number of such sequences is equivalent to the number of functions from $\mathbb{N}$ to itself, which is provably uncountable.

$q \pm \frac{1}n$ is a sequence converging to $q$ for any sequence of $\pm$ signs.

That is a lower bound matching the stated upper bound. This solves the problem in the sense that the Schroeder Bernstein principle applies. Finding a specific 1-1 correspondence between rational convergent sequences and 0-1 sequences is more complicated.

We have that non-repeating (injective) sequences of elements in $\{\,1+1/n:n\in\mathbb{N}\,\}$ form a continuum, and all of them have limit $1$, so our set is at least a continuum. Since also all rational sequences form a continuum, our set is also at most a continuum.