Suppose that $a_n\neq0$ for all $n\in\mathbb{N}$ and $L=\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|$. Prove that $a_{n}\to 0$. If $L<1$, how do I prove that $\lim_{n\rightarrow\infty}a_n=0$ ?
I tried setting $L$ as $1-\delta$, because
$$L<1\Longleftrightarrow L=1-\delta,\, \delta>0$$
So, by definition:
$$\forall\varepsilon>0, \exists N>0:n>N\Rightarrow\left|\left|\frac{a_{n+1}}{a_n}\right|-1+\delta\right|<\varepsilon$$
But I don't think that solving that inequality will help me at all, so I got stuck.

Let $\epsilon > 0$. Let $N$ be large enough so that $\big| \frac {a_{n+1}} {a_n} \big| < 1 - \epsilon$ for all $n \ge N$. Then,
$$
\frac {1} {a_n} =
\frac {a_{n+1}} {a_n} \cdot
\frac {a_{n+2}} {a_{n+1}} \cdot
\frac {a_{n+3}} {a_{n+2}} \cdot
\frac {a_{n+4}} {a_{n+3}} \cdots
\frac {a_{n+k}} {a_{n+k-1}} \cdot
\frac {1} {a_{n+k}} \iff \\
\frac {a_{n+k}} {a_n} =
\frac {a_{n+1}} {a_n} \cdot
\frac {a_{n+2}} {a_{n+1}} \cdot
\frac {a_{n+3}} {a_{n+2}} \cdot
\frac {a_{n+4}} {a_{n+3}} \cdots
\frac {a_{n+k}} {a_{n+k-1}}
$$
So that,
$$
\Big| \frac {a_{n+k}} {a_n} \Big| =
\Big| \frac {a_{n+1}} {a_n} \cdot
\frac {a_{n+2}} {a_{n+1}} \cdot
\frac {a_{n+3}} {a_{n+2}} \cdot
\frac {a_{n+4}} {a_{n+3}} \cdots
\frac {a_{n+k}} {a_{n+k-1}} \Big| < \\
(1-\epsilon)^k \xrightarrow{k \rightarrow \infty} 0 \iff \\
a_{n+k} \xrightarrow{k \rightarrow \infty} 0
$$
I.e., $\lim_{n \rightarrow \infty} a_n = 0$.
