Problem:
If $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 0$, show that $\lim_{n\to\infty}a_n = 0$.
Attempted Proof:
Let $\epsilon > 0$. From the hypothesis, $\exists \ N \in \mathbb{P}$ such that if $n \geq N$, then $$\left|\ \frac{a_{n+1}}{a_n} - 0 \ \right| < \epsilon.$$
This implies
$$\left| {a_{n+1}} \right| < \epsilon \left| a_n \right|.$$
Thus, let $n\geq N'$ such that $$\left|\ \frac{a_{n+1}}{a_n} - 0 \ \right| < \epsilon\left|a_n\right|.$$
Then we have $\left|a_{n+1}\right| < \epsilon,$ which implies $\lim_{n\to\infty}a_n = 0$.
My main concern with my proof is that $\epsilon$ depends on $a_n$. Is this an issue?