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.
 A: 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!
A: 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., $\lim_{n \rightarrow \infty} a_n = 0$.
A: $L:=\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$
Choose $L<r<1 $ . Then $\exists N\in \mathbb{N} $ such that $\forall n\ge N $ $$\left|\dfrac{a_{n+1}}{a_n}\right|<r$$
$\implies |a_{n+1}|<r|a_n| \space, \space \forall n\ge N$
Now, we have
$\begin{align*}\left| {{a_{N + 1}}} \right| & < r\left| {{a_N}} \right|\\ \left| {{a_{N + 2}}} \right| & < r\left| {{a_{N + 1}}} \right| < {r^2}\left| {{a_N}} \right|\\ \left| {{a_{N + 3}}} \right| & < r\left| {{a_{N + 2}}} \right| < {r^3}\left| {{a_N}} \right|\\ & \hspace{0.5in} \vdots \\ \left| {{a_{N + k}}} \right| & < r\left| {{a_{N + k - 1}}} \right| < {r^k}\left| {{a_N}} \right|\end{align*}$
Hence for $k\in\Bbb{N}$ ,$ \left| a_{N+k} \right | <r^k |a_N|$
$0\le\lim_{k\to \infty} |a_{N+k}| \le |a_N| \lim_{k\to\infty} r^k$
$\implies \lim_{k\to\infty} |a_{N+k}|=0$
Hence $(a_n) $ converges to $0$.(Proved)
