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.

Lemma
Let $A\in\textbf{R}$ and $B\in\textbf{R}$ where $A < B$. If the real-valued sequences $(x_{n})_{n\geq 1}$ and $(y_{n})_{n\geq 1}$ converges to $A$ and $B$, respectively, then there exists some $N\in\textbf{N}$ such that $x_{n} < y_{n}$ for $n\geq N$.
Proof
Let $C\in\textbf{R}$ be such that $A < C < B$.
Then for $\varepsilon = C - A$ there corresponds $n_{0}\in\textbf{N}$ s.t. for every $n\in\textbf{N}$
\begin{align*}
n\geq n_{0} \Rightarrow |x_{n} - A| < \varepsilon = C - A \Rightarrow x_{n} - A < C - A \Rightarrow x_{n} < C
\end{align*}
On ther other hand, let $\eta = B - C$. Then there corresponds $n_{1}\in\textbf{N}$ s.t. for every $n\in\textbf{N}$
\begin{align*}
n\geq n_{1} \Rightarrow |y_{n} - B| < \eta = B - C \Rightarrow B - y_{n} < B - C \Rightarrow y_{n} > C
\end{align*}
Gathering both results, we may claim that $x_{n} < C < y_{n}$ for every $n\geq N$, where $N = \max\{n_{0},n_{1}\}$.
Hence the target lemma is true.
Solution to the exercise
We may now prove the proposed claim. Since $L < 1$, there exists $r\in\textbf{R}$ such that $L < r < 1$.
Consequently, there exists $N\in\textbf{N}$ so that $\left|\dfrac{a_{n+1}}{a_{n}}\right| < r$ for $n\geq N$.
Hence one may conclude that:
\begin{align*}
|a_{n+1}| = \left|\frac{a_{n+1}}{a_{n}}\times\frac{a_{n}}{a_{n-1}}\times\ldots\times\frac{a_{N+1}}{a_{N}}\times a_{N}\right| < |a_{N}|r^{n - N + 1}
\end{align*}
where the RHS is a geometric sequence whose ratio $r\in(0,1)$.
Taking the limit from both sides, one deduces that
\begin{align*}
\lim_{n\to\infty}|a_{n+1}| \leq \lim_{n\to\infty}|a_{N}|r^{n-N +1} = 0,
\end{align*}
and we are done.
Hopefully this helps!
