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

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

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