$\lim a_n = 0$ if $\left|\frac{a_{n +1}}{a_n}\right|\to L < 1$ Suppose that $a_n \neq 0$, for every n and that $L = \lim |\frac{a_{n +1}}{a_n}|$ exists. Show that if L < 1, then $\lim a_n = 0$.
What I did so far:
If L < 1 and $L = \lim |\frac{a_{n +1}}{a_n}|$, there exists $n_0$ such that for $n \geq n_0$, $0 < |a_{n+1}| < |a_n|$. That means that the sequence $(|a_n|)_{n \geq n_0}$ is decreasing. 
Consider the set $S = \{ |a_0|, ..., |a_{n_0} \}$. S is finite. Let $\beta = \max_{0 \leq i \leq n_0} |a_ i|$. Thus, $(|a_n|)$ is limited (because, for every n, $0 \leq |a_n| \leq \beta).$
Now I have that $(|a_n|)$ is limited and, throwing away a finite number of terms (the $n_0$ firsts) I can assume that it is decreasing. So I know that $(|a_n|) converges. 
How can I prove that it converges to $0$? 
I also know that if I prove that $ \lim |a_n| = 0$, then I have that $ \lim a_n = 0$.
 A: You showed that the sequence converges, and it's a good start. But there is an easier way : take a real number $\alpha <1$ such that $\frac{|a_{n+1}|}{|a_n|}< \alpha < 1$ after a certain rank, which is possible because the limit of $\frac{|a_{n+1}|}{|a_n|}$ is strictly les than $1$. Then, $$|a_{n+1}|<\alpha |a_n|$$
$$|a_{n+2}| < \alpha |a_{n+1}|< \alpha^2 |a_n|$$
$$|a_{n+3}|<...<\alpha^3 |a_n|$$
and so on.
Now do you see why the sequence is convergent to $0$ ?
A: Since $L < 1$, you know that there is some $n_0$ such that for all $n > n_0$ you have $\vert a_{n+1}/a_n \vert < 1 - \delta$ for $\delta = (1 - L)/2$, say. Note that $1-\delta \in (0,1)$.
Then you have for all $n > n_0$ that $a_n < (1-\delta)^{n - n_0} a_{n_0}$.
This reduces the problem to showing that $\lim_{ n \to \infty} x^n = 0$ for $x \in (0,1)$, which I suspect you can handle.
A: Others have given a direct answer. Nevertheless, if you want to continue on what you have built, you can proceed as such:
You have proven that $(|a_n|)$ converges. Let $\lambda$ be its limit. If $\lambda \neq 0$, we obtain $$\lim_{n\to +\infty}\frac{|a_{n+1}|}{|a_n|} = \frac \lambda \lambda = 1$$which contradicts your assumption. Therefore, $\lambda = 0$, as required.
A: If $$\lim_{n\to\infty }\left|\frac{a_{n+1}}{a_n}\right|=\ell<1$$
then there is a $N$ such that $$\left|\frac{a_{n+1}}{a_n}\right|\leq \frac{1+\ell}{2}$$
if $n\geq N$ and thus, if $n\geq N$
$$|a_{n+1}|\leq \left(\frac{1+\ell}{2}\right)^{n-N}|a_{N}|\underset{n\to\infty }{\longrightarrow }0$$
since $0\leq\frac{1+\ell}{2}<1$.
