Since the number of positive divisors of a positive integer $m$ does not exceed $2\sqrt{m}$, and $a_1, \dots, a_k$ are pairwise distinct positive divisors of $[a_1, \dots, a_k]$, it follows that $[a_1, \dots, a_k] \ge k^2/4$. Consequently,
$$\begin{aligned} \sum_{k=1}^{n} \frac{1}{[a_1, \dots, a_k]} &= \frac{1}{a_1} + \sum_{k=2}^{n} \frac{1}{[a_1, \dots, a_k]} \le 1 + \sum_{k=2}^{n} \frac{4}{k^2} < 1 + 4 \sum_{k=2}^{n} \frac{1}{k^2 - \frac{1}{4}} \\ &= 1 + 4 \cdot 2 \left( \frac{1}{3} - \frac{1}{2n+1} \right) < \frac{11}{3} < 4. \end{aligned}$$