Prove: lim|$\frac{a_{n+1}}{a_n}$| = 0 $\implies$ lim $a_n$ = 0. (Show |$a_{n+1}$| < $\frac {1}{2}$ |$a_n$| for n $\gg$ 1.) The book I am using for my Advance Calculus course is Introduction to Analysis by Arthur Mattuck.
Prove: $\lim\left|\frac{a_{n+1} }{a_n}\right| = 0 \implies \lim a_n = 0$. (Show $\left|a_{n+1} \right| < \frac {1}{2} \left|a_n\right| \text{ for } n \gg 1$.)
This is my rough proof to this question. I was wondering if anybody can look over it and see if I made a mistake or if there is a simpler way of doing this problem. I want to thank you ahead of time it is greatly appreciated.So lets begin:
Proof:
Since $|l|<1$, therefore chose $\varepsilon_0>0$ such that $|l|+\varepsilon_0=h<1\qquad (1)$.
Since $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=l$, therefore there exists a positive integer $m$ such that $$\left|\dfrac{a_{n+1}}{a_n}-l\right|<\varepsilon_0\quad\text{for all }n\geq m.$$ Now, $$\begin{align}\left|\dfrac{a_{n+1}}{a_n}\right|&=\left|\left(\dfrac{a_{n+1}}{a_n}-l\right)+l\right|\leq\left|\dfrac{a_{n+1}}{a_n}-l\right|+|l| \\ & <\varepsilon_0+|l|\quad\text{for all }n\geq m\\ &\implies\left|\dfrac{a_{n+1}}{a_n}\right|<h\quad\text{for all }n\geq m \end{align}$$
Replacing $n$ by $m,m+1,m+2,\ldots,n-1$. 
And also multiply the corresponding sides of the resulting $(n-m)$ inequalities, we have 
$$\begin{align}\implies & \left|\dfrac{a_{m+1}}{a_m}\right|\cdot\left|\dfrac{a_{m+2}}{a_{m+1}}\right|\cdot\left|\dfrac{a_{m+3}}{a_{m+2}}\right|\cdots\left|\dfrac{a_n}{a_{n-1}}\right|<h^{n-m} \\ \implies & \left|\dfrac{a_{m+1}}{a_m}\cdot\dfrac{a_{m+2}}{a_{m+1}}\cdot\dfrac{a_{m+3}}{a_{m+2}}\cdots\dfrac{a_n}{a_{n-1}}\right|<h^{n-m} \\ \implies & \left|\dfrac{a_n}{a_m}\right|<h^{n-m} \\ \implies & \left|\dfrac{a_n}{a_m}\right|<\dfrac{h^n}{h^m} \\ \implies & \dfrac{\left|a_n\right|}{\left|a_m\right|}<\dfrac{h^n}{h^m} \\ \implies & \left|a_n\right|<\dfrac{h^n}{h^m}\left|a_m\right|\quad\text{for all }n\geq m \end{align}$$ Also, since $0<h<1$, therefore, $h^n\to 0$.
Thus given for $\varepsilon>0$, there exists a positive integer $p$ such that $$\left|h^n\right|<\dfrac{h^m\varepsilon}{\left|a_m\right|}\quad\text{for all }n\geq p\tag{2}$$
Let $q=\max\{m,p\}$ then $$\left|a_n\right|<h^n\dfrac{\left|a_m\right|}{h^m}<\dfrac{h^m\varepsilon}{\left|a_m\right|}\cdot\dfrac{\left|a_m\right|}{h^m}=\varepsilon\quad\text{for all }n\geq q.$$ That implies $\left|a_n\right|<\varepsilon\quad\text{for all }n\geq q$.
That means $\lim\limits_{n\to\infty}a_n=0$.
 A: Hint:
Simpler way may be the following, which is a soft one; you can easily extend it to the degree of rigor you want:
If $|a_{n+1}/a_{n}| \to 0$ as $n$ grows indefinitely, then there is some $N \geq 1$ such that $|a_{n+1}| < |a_{n}|/2$ for all $n \geq N$, so $|a_{N+k}| < |a_{N}|/2^{k}$ for all $k \geq 1$, and hence $|a_{N+k}| \to 0$ as $k$ grows indefinitely.
A: Since $\lim|\frac{a_{n+1}}{a_n}| = 0$ we have that for every $\epsilon > 0$ there exists an $ n_0 \in \mathbb{N}$ such that $|\frac{a_{n+1}}{a_n}| < \epsilon$ when evr $n\geq n_0$
If $\epsilon = 1/2$ then we get $|a_{n+1}| < \frac{1}{2}|a_n|$ for all $n\geq k$ which we can choose to be the smallest such $k$. 
applying this repeatedly till the $k$th stage we get $0 \leq |a_{n+1}| < \frac{1}{2^{n-k} }|a_k|= \frac {M} {2^n} $ where $M$ is the finite number $2^k.|a_k|$.Then by squeeze principal we get $\lim a_n = 0$
I think this is simpler!
A: Depending by what you covered, here is an alternate way to solve the problem.
Since $\lim_n \left| \frac{a_{n+1}}{a_n} \right| =0$ there exists an $N$ s that 
$$\left| \frac{a_{n+1}}{a_n} \right| <1 \forall n >N$$
This shows that $|a_n|$ is eventually decreasing. As it is also bounded from below (by 0), $|a_n|$ is convergent.
If $l=\lim_n |a_n|$ then we have $l \geq 0$. 
We cannot have $l >0$ because then
$$0=\left| \frac{a_{n+1}}{a_n} \right| =\frac{l}{l} =1 \,.$$
Therefore $l=0$. 
Finally, since $\lim_n |a_n| =0$ we get immediately that $\lim_n a_n =0$ .
P.S> This proof is a bit of overkill.
