real analysis - $lim a_n = 0$ 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 $\lim a_n = 0$.