Romanian Mathematical Olympiad

For $m=n\in {\mathbb{N}}^{\ast }$ we get $a_n\le 1$, implying that the sequence is upper bounded.

Denote by $L=\sup \{a_n\mid n\in {\mathbb{N}}^{\ast }\}$. We get $L\le 1$. If $\epsilon >0$ is arbitrary we find $n_0\in {\mathbb{N}}^{\ast }$ such that $L-\epsilon <a_{n_0}\le L$.

Consider $m_0=\max \{n_0,1+\lfloor \frac{1}{a_{n_0}-L+\epsilon }\rfloor \}$. For $m\ge m_0$ we have $$a_m\ge a_{n_0}-\frac{1}{m}\ge a_{n_0}-\frac{1}{m_0}>L-\epsilon ,$$ That is $|a_m-L|<\epsilon$ for all $m\ge m_0$ proving that the sequence has $L$ as limit.