Romanian Mathematical Olympiad

For $k \in \mathbb{N}^*$, the sequence $(y_n^{(k)})_{n \ge 1}$ is convergent and its terms are integers. Then there exist $n_k, a_k \in \mathbb{N}^*$ so that $y_n^{(k)} = a_k, \forall n \ge n_k$. Consequently, $x_n^k \in [a_k, a_k + 1), \forall n \ge n_k$. In particular, $x_n \in [a_1, a_1 + 1), \forall n \ge n_1$. It follows that the sequence $(x_n)_{n \ge 1}$ is bounded.

Suppose, by contradiction, that the sequence $(x_n)_{n \ge 1}$ has two limit points $a$ and $b$, with $1 \le a < b$. Then there exist two subsequences $(x_{i_n})_{n \ge 1}$ and $(x_{j_n})_{n \ge 1}$ of $(x_n)_{n \ge 1}$ such that $\lim_{n \to \infty} x_{i_n} = a$ and $\lim_{n \to \infty} x_{j_n} = b$. Let $k \in \mathbb{N}^*$. From $i_n, j_n \ge n$, $\forall n \in \mathbb{N}^*$, we get $x_{i_n}^k, x_{j_n}^k \in [a_k, a_k + 1), \forall n \ge n_k$. It follows $x_{j_n}^k - x_{i_n}^k < 1, \forall n \ge n_k$. Taking the limit for $n \to \infty$ gives $b^k - a^k \le 1$. Therefore $b^k - a^k \le 1, \forall k \in \mathbb{N}^*$. On the other hand, $1 \le a < b$ implies $\lim_{k \to \infty} (b^k - a^k) = \infty$, in contradiction with the previous inequality.

This shows that the sequence $(x_n)_{n \ge 1}$ is convergent.