If $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 0$, show that $\lim_{n\to\infty}{a_n} =0$. 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?
 A: Since $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=0$, then for $\epsilon=\frac12$, there is $N\in\mathbb{N}$ such that when $n\ge N$,
$$ \bigg|\frac{a_{n+1}}{a_n}\bigg|<\epsilon. $$
Thus for $n>N$, one has
$$ \bigg|\frac{a_{N+1}}{a_N}\bigg|<\epsilon, \bigg|\frac{a_{N+2}}{a_{N+1}}\bigg|<\epsilon,\cdots,\bigg|\frac{a_{n}}{a_{n-1}}\bigg|<\epsilon$$
and hence
$$ \bigg|\frac{a_{n}}{a_N}\bigg|<\epsilon^{n-N}$$
or
$$ |a_n|<\epsilon^{n-N}|a_N|. $$
So
$$ \lim_{n\to\infty}a_n=0.$$
A: For $\epsilon =1$, one gets $|a_{n+1}| < |a_n|$ for each $n \ge N$, so as $(|a_n|)$ is bounded below, we deduce that it is convergent. Suppose that it converges to some $L >0$ and get a contradiction by proving that $\lim \frac{a_{n+1}}{a_n}$ would then be $1$. Therefore it converges to $0$ and so does $(a_n)$.
A: That would be an issue, yes. The quickest way to prove this is to recognize that $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=0$ tells you that $\sum_0^\infty a_n$ converges by the ratio test. Thus $a_n$ must go to $0$. 
A: You should study $|a_n|$ and see that after $N$ big enough, you have $|a_{n+1}|<|a_n|$...
