Let $a_n \neq 0 \;\; \forall n \in \mathbb{N}$ and $\exists L = \lim |\frac{a_{n+1}}{a_n}|$. Show that, if $L<1$, then $\lim a_n = 0$

Question

Let $a_n \neq 0 \;\; \forall n \in \mathbb{N}$ and $\exists L = \lim |\frac{a_{n+1}}{a_n}|$. Show that, if $L<1$, then $\lim a_n = 0$

Please verify if my attempt makes sense and how could I fix it. I also would appreciate to see another ways of solving this exercise.

Choose $c$ such that $L<c<1$. Then $\exists n_1 \in \mathbb{N}$ such that $n \geq n_1 \implies |\frac{a_{n+1}}{a_n}|<c$ (already verified this)

$\implies|a_{n+1}| < c|a_n| \implies\frac{|a_{n+1}|}{c^{n+1}} < \frac{|a_n|}{c^n}$

Then $\frac{|a_n|}{c^n}$ is monotonically decreasing for $n \geq n_0$.

Since $\frac{|a_n|}{c^n} \geq 0$, this sequence is limited, then there is $n_1\geq n_0$ such that $\exists M>0$ such that $0 \leq \frac{|a_n|}{c^n} \leq M \implies 0 \leq |a_n| < Mc^n$. By squeeze theorem, $\lim |a_n| = 0$.

But I'm failing to conclude $\lim a_n = 0$, all I can see is that $\lim a_n \leq 0$.

Answer 1

Your proof is correct and you can conclude that $\lim\limits_{n\to\infty} a_n=0$ from the inequalities

$$-\vert a_n\vert\le a_n\le \vert a_n\vert$$

Answer 2

What you did is correct. I don't understand why is it that you say that you only proved that $\lim_{n\to\infty}a_n\leqslant0$. You proved that $\lim_{n\to\infty}\lvert a_n\rvert=0$ and this is equivalent to asserting that $\lim_{n\to\infty}a_n=0$.

Another way of proving this consists in taking $c$ as your $c$. So $\left\lvert\frac{a_{n+1}}{a_n}\right\rvert<c$, which means that $\lvert a_{n+1}\rvert\leqslant c\lvert a_n\rvert$. So, $\lvert a_n\rvert\leqslant c^n\lvert a_0\rvert$ and therefore the series $\sum_{n=0}^\infty\lvert a_n\rvert$ converges, which implies that $\lim_{n\to\infty}\lvert a_n\rvert=0$.