Suppose $a_n > 0, b_n > 0$ an $\lim \limits_{n \to \infty} \frac{a_n}{b_n} = L$ with $L_1 \in \mathbb{R}_+$. Prove that if $\sum_{n \in \mathbb N} a_n$ converges, then so does $\sum_{n \in \Bbb N} b_n$.
Suppose $\sum_{n \in \Bbb N} a_n$ converges to $L_a$. Since $\lim \limits_{n \to \infty} \frac{a_n}{b_n} = L_1$, we have $a_n = b_n(L_1 + \varepsilon(n))$ with $\varepsilon(n) \to 0$ as $n \to \infty$. Therefore, $\sum_{n \in \Bbb N} a_n = \sum_{n \in \Bbb N} b_n(L_1 + \varepsilon(n)) = L_a$, so $\lim \limits_{n \to \infty} \sum_{i=1}^n b_n(L_1 + \varepsilon(n)) = L_a$, and therefore, $\sum…