Question 65
For two rational numbers $r, s$ we say $r \mid s$ if there is $k \in \mathbb{Z}$ so that $s = kr$. The sequence $(a_n)_{n \in \mathbb{N}}$ is an increasing sequence of natural numbers such that for all $i \neq j \in \mathbb{N}$, $\gcd(a_i, a_j) = 1$, and $(b_n)_{n \in \mathbb{N}}$ is a sequence of distinct natural numbers. Assume that for each $n \in \mathbb{N}$ we have
$$
\sum_{i=1}^{n} \frac{1}{a_i} \mid \sum_{i=1}^{n} \frac{1}{b_i},
$$
prove that for all $n \in \mathbb{N}$ we have $a_n = b_n$.
$$
\sum_{i=1}^{n} \frac{1}{a_i} \mid \sum_{i=1}^{n} \frac{1}{b_i},
$$
prove that for all $n \in \mathbb{N}$ we have $a_n = b_n$.