Let $(a_n)_{n \geq 1} \subset \mathbb{R}$ be a monotonic and convergent sequence, such that for all monotonic and convergent sequences $(b_n)_{n \geq 1}$ we have that $$(a_n+b_n), \: (a_n-b_n), \: (a_nb_n)$$ are also monotonic and convergent. Prove that $(a_n)$ is eventually constant.
I tried to use a proof by contradiction. Say $a_n$ is increasing and convergent to $\alpha$. If the claim isn't true, then $\forall n \geq 1$ there is a minimal number $k_n>n$ such that $a_{k_n}>a_n$ and so $\alpha>a_{k_{n+1}}\geq a_{k_n}$. Then, I tried to use the given property, setting $b_n=a_{k_n}$ and get a contradiction, but I can't see it...
Let $b_n$ be constant on some stretches. If $(a_n)$ is not eventually constant, there is a subsequence $(a_{k_m})$ that is strictly monotonic. For $k_{2m} \leqslant n < k_{2m+2}$, let $b_n = a_{k_{2m+1}}$. Then $(b_n)$ is monotonic with the same limit as $(a_n)$. The sequence $(a_n - b_n)$ converges to $0$ and $a_{k_{2m+1}} - b_{k_{2m+1}} = 0$ for all $m$, but $a_{k_{2m}} - b_{k_{2m}} = a_{k_{2m}} - a_{k_{2m+1}} \neq 0$, so $(a_n - b_n)$ is not monotonic.