My question is perhaps elementary, still I'm neither managing to figure out counterexamples nor to prove the result.
Let $(x_n)_{n\geq 1}$ be a real-valued sequence and assume that $x_n\to x\in \mathbb{R}$. Is it true that there exists a monotonic decreasing positive sequence $(r_n)_{n \geq1}$ such that $r_n\downarrow 0$ and $|x_n-x|<r_n$ ultimately? Obviously, the question can be extended to convergence on metric spaces, questioning whether $d(x_n,x)<r_n$ ultimately, for the given distance $d$.
EDIT It seems to be always possible to construct a monotone nonincreasing sequence, by simply reasoning as follows: let $\epsilon_1>\epsilon_2>\epsilon_3> \ldots$ be any positive decreasing sequence of reals, decaying to $0$. Then:
1. $\exists n_1 \in \mathbb{N}$, s.t. $\forall n \geq n_1$, $d(x,x_n)<\epsilon_1$;
2. $\exists n_2 \in \mathbb{N}$, s.t. $\forall n \geq \max(n_1,n_2)$, $d(x,x_n)<\epsilon_2$;
$\vdots$
i. $\exists n_i \in \mathbb{N}$, s.t. $\forall n \geq \max(n_1, \ldots, n_i)$, $d(x,x_n)<\epsilon_i$;
$\vdots$
Thus we can set $r_n= \epsilon_i$, $\forall n \in [\max(n_1, \ldots,n_{i}),\max(n_1, \ldots,n_{i+1})) \cap \mathbb{N}$. This would give a sequence such that $r_n \downarrow 0$ and $r_n > d(x_n,x)$ ultimately.
