I'm trying a proof technique I'm not used to for limits on fractions, which attempts to avoid an epsilon-delta approach similarly to how the single variable chain rule is proved in baby Rudin, and I was wondering if it works. Any help or tips are very welcome!
Statement: If $a_n, b_n>0$ and $\lim \limits_{n \to \infty} \frac{a_n}{b_n} = L_1$ with $L_1 > 0$, then if $\sum_{n \in \mathbb N} a_n$ converges, so does $\sum_{n \in \Bbb N} b_n$
Proof: 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 $$\lim \limits_{n \to \infty} \sum_{i=1}^n b_i(L_1 + \varepsilon(n)) = L_a,$$
and therefore,
$$\sum_{i=1}^nb_i=\frac{L_a+\mu(n)}{L_1 + \varepsilon(n) } ,$$ where $\mu(n) \to 0$ as $n \to \infty$.
Therefore, letting $n \to \infty$ results in $\sum_{i=1}^nb_i = \frac{L_a}{L_1}$, so $\sum_{n \in \mathbb N}b_n$ converges. $\blacksquare$
Best Answer
Actually, you should have $$\lim_{n\to\infty}\sum_{i=1}^n b_i(L_1+\epsilon(\color{red}{i}))=L_a$$
(note the red colored $i$ inside the sum).
Also, even if the continuation would be correct, you are jumping over a lot of steps.
This is a big leap. I don't see where you get the latter expression from the former. You really should expand this out to explain exactly how the limit implies that the sum is equal to $\frac{L_a+\mu(n)}{L_1 + \varepsilon(n)}$, and explain exactly what $\mu(n)$ is.