If a sequence converges, then if its quotient converges, then it converges to less than or equal to one I want to show the following.

Let $a_n$ converge. If $\frac{a_{n+1}}{a_n}$ converges then $|\lim\frac{a_{n+1}}{a_n}|\leq 1$.

I attach my proof below, but I would be happy to see if there is a more elegant and simple one. I would think one could use the following theorem, but I do not know how:

If $a_n$ converges and $\frac{a_{n+1}}{a_n}$ converges, then $\sqrt[n]{a_n}$ converges and $\lim\left(\frac{a_{n+1}}{a_n}\right)=\lim\left(\sqrt[n]{a_n}\right).$

My proof - mind that the students have only learned the definitions regarding convergence, basic theorems regarding convergence and convergence to infinity. They still haven't learned the theorems regarding convergence and monotonic series, partial limits, Cauchy series, etc.
Let us assume that $a_n$ converges and $\frac{a_{n+1}}{a_n}$ converges.
Let us denote $\lim(a_n)=L$.
$$\lim(a_{n+1})=\lim(\frac{a_{n+1}}{a_n}\cdot a_n)=\lim(\frac{a_{n+1}}{a_n})\lim(a_n).$$
$$L=\lim(\frac{a_{n+1}}{a_n})L,$$
$$L(1-\lim(\frac{a_{n+1}}{a_n}))=0,$$
$$L=0 \vee \lim(\frac{a_{n+1}}{a_n})=1.$$
The second case - proves our goal instantly.
Let us assume that $L=0$. Let us assume in negation that $\lim(\frac{a_{n+1}}{a_n})>1$ (we would also need to prove the other case).
Let us denote $\lim(\frac{a_{n+1}}{a_n})=D$.
Because the limit of $\lim(\frac{a_{n+1}}{a_n})$ exists then:
$$\forall \epsilon >0. \exists n_0.\forall n>n_0 . \lim(\frac{a_{n+1}}{a_n}) \in (d-\epsilon,D+\epsilon).$$
Specifically for $\epsilon=\frac{D-1}{2}$. So:
$\exists n_0.\forall n>n_0. D-\frac{D-1}{2}<\frac{a_{n+1}}{a_n}$
$\exists n_0.\forall n>n_0. \frac{D+1}{2}a_n<a_{n+1}$
Now because $a_n$ converges to 0 then:
$\forall \epsilon>0.\exists n_0. \forall n>n_0. a_n\in (-\epsilon,\epsilon)$
But because $\exists n_0.\forall n>n_0. \frac{D+1}{2}a_n<a_{n+1}$ for some $\epsilon$ for all $n_0$ we could find a $n>n_0$ that $a_n\notin (-\epsilon,\epsilon)$ in negation with the assumption that $a_n$ converges.
 A: 
If a sequence convereges, then if it's quotient converges, then it converges to less than or equal to one.

I am interpreting the assertion to be that $\langle a_n\rangle$ is a Real valued sequence and that if $~\displaystyle \frac{a_{n+1}}{a_n}~$ converges to a value $L$, where $|L| > 1$, then the sequence $\langle a_n\rangle$ is not convergent.

If $|L| > 1,$ then $~\displaystyle \lim_{n \to \infty} |L|^n = \infty.$
Assume that $|L| > 1,$ with the quotient converging to $L$.
Without loss of generality, after a finite number of terms in the sequence, every term thereafter must be non-zero.  Otherwise, if zero terms (occasionally) occurred ad infinitum, you could not have the quotient converging to a value $L$ such that $|L| > 1.$
Then, after a finite number of terms, the quotient will be within (for example) $L \pm \epsilon$, where $\epsilon$ can be made as small as desired.
So, you can choose $\epsilon$ small enough so that after a finite number of terms, the absolute value of the quotient will be greater than $|L - \epsilon| > 1.$
So, assume that for all $n \geq N$, $~\displaystyle \left|\frac{a_{n+1}}{a_n}\right|~ > |L - \epsilon| > 1.$ 
Further, it is being assumed that $a_N \neq 0$.
Then, you will have that for all $n > N$,
$~\displaystyle |a_n| > |a_N| \times |L - \epsilon|^{n - N}.$
This means that as $n \to \infty, ~|a_n|~$ will grow unbounded.
This means that $\langle a_n\rangle$ can not be a convergent sequence.
A: We may assume that $a_n\neq 0$ for any integer $n.$
If $\underset{n\to \infty}\lim a_n=\ell\neq 0$, then the limit of the quotient is $\frac{\ell}{\ell}=1.$
If $\ell=0,$ then since $-|a_n|\le a_n\le |a_n|$, we may assume $a_n>0.$
Suppose that for each integer $k$ there is an integer $n_k$ such that $\frac{a_{n_k+1}}{a_{n_k}}>1$ and $n_{k+1}>n_k+1.$ Then, $\underset{n\to \infty}{\lim}\frac{a_{n+1}}{a_n}\ge 1.$ (It is here we use the fact that this limit exists.) But in this case, $a_n \nrightarrow 0.$ Therefore, there is some integer $k$ for which no such $n_k$ exists. That is, the sequence $(a_{n_k})$ is finite or empty. It follows that $\frac{a_{n+1}}{a_n}\le 1$ if $n$ is large enough.
Remark: the requirement that the limit of the quotient exists is necessary. For example, the sequence $\{1,1/2,1/4,1/3/1/8,1/4,1/16,1/5,\cdots\}$ converges to zero and yet $\underset{n\to \infty}\limsup \frac{a_{n+1}}{a_n}=\infty$ and $\underset{n\to\infty}\liminf \frac{a_{n+1}}{a_n}=0.$
