Proof Verification of a convergence problem I am self studying and the solutions to the book I am using is not given. Thus, I want to make sure that this is correct.
Question: Suppose $(a_{n})$ is a sequence in $\mathbb R$ such that $\lim_{n\rightarrow \infty} \vert{\frac{a_{n+1}}{a_n}}\vert=c<1$. Show that $a_n \rightarrow 0$
Answer: Since $c<1$, $\exists n_0\in \mathbb N$ such that for all $n \geq n_0$, $\vert{\frac{a_{n+1}}{a_n}}\vert<1$ or $|a_{n+1}|< |a_n|$. Let us take $b_n=|a_{n_0+n}|$ for all $n \in \mathbb N$. $(b_n)$ is $n_0-\rm tail$ of $(|a_n|)$ and it is a monotonically decreasing sequence bounded below by $0$. Thus, it must be convergent and thus, $(|a_n|)$ must be convergent and $c_n=|a_{n+1}|$ converges to the same value as $(|a_n|)$. Now,if $|a_n|  \rightarrow a \neq 0$,then, we immediately get that $c=\lim_{n\rightarrow \infty} \vert{\frac{a_{n+1}}{a_n}}\vert=1$ by algebra of limits. This contradicts the hypothesis in question. Thus,the only possiblity is $a=0$. $|a_n|\rightarrow 0$ and thus, $a_n \rightarrow 0$.
 A: Your proof is fine, as an alternative approach we can use that since eventually for $n\ge n_0$
$$\left|{\frac{a_{n+1}}{a_n}}\right|\le\frac{c+1}2=k<1$$
we have that
$$|a_n|\le k^{n-n_0}|a_{n_0}| =\frac{|a_{n_0}|}{k^{n_0}}k^n\to 0$$
See also the related

*

*proof that $a_n$ is a null sequence
A: With OP's question answered by subsequent comment, it is open season on alternative approach.
$\underline{\text{Lemma 1}}$ 
Given $0 < c < 1, ~0 < \epsilon < 1$ there exists $N \in \Bbb{Z^+}$ 
such that $\forall n \in \Bbb{Z^+}$ such that $n \geq N, c^n < \epsilon.$
Proof:
Choose  $\displaystyle N \in \Bbb{Z^+}$ such that
$N > \frac{\log(\epsilon)}{\log(c)}.$ 
Then, $N \log(c) < \log \epsilon \implies c^N < \epsilon.$
Further, $~n > N \implies n\log(c) < N \log(c) < \log \epsilon \implies c^n < \epsilon.$

From the constraint, you must have that none of $a_1, a_2, a_3 \cdots$ are $= 0.$
$$|a_{n+1}| = |a_1| \times \prod_{k=1}^n \left|\frac{a_{k+1}}{a_k}\right| = |a_1| \times c^n.\tag1 $$
By Lemma 1, $\lim_{n\to\infty} c^n = 0.$ 
Therefore, since $a_1$ is a finite number, (1) above
implies that
$$\lim_{n\to\infty} |a_n| = 0.$$
This implies that $\displaystyle \lim_{n\to\infty} a_n = 0.$
