Show that $a_n$ converges using Cauchy

Question

Let $a_n$ be a sequence and let $0<q<1$ be a constant. Show that $a_n$ converges given that $\forall n\in \mathbb{N}:(n\geq 2\implies|a_{n+1}-a_n|<q|a_n-a_{n-1}|).$

progress.
We'll prove it using Cauchy's condition: $$(\forall\epsilon>0)\:(\exists N\in\mathbb{N})\:(\forall n\in\mathbb{N})\:(\forall p\in\mathbb{N})\:(n>N\implies|a_{n+p}-a_n|<\epsilon)$$we have
$\eqalign{\lvert a_{n+p}-a_n\rvert&=\lvert a_{n+p}-a_{n+p-1}+a_{n+p-1}-...-a_{n+1}+a_{n+1}-a_n\rvert \\ &\le\lvert a_{n+p}-a_{n+p-1}\rvert+|a_{n+p-1}-a_{n+p-2}\rvert+...+\lvert a_{n+1}-a_n\rvert \\ &\le q^p\lvert a_n-a_{n-1}\rvert+q^{p-1}\lvert a_n-a_{n-1}\rvert+...+\lvert a_n-a_{n-1}\rvert \\ &\le\lvert a_n-a_{n-1}\rvert(q^p+q^{p-1}+...+1) \\ &\le p\lvert a_n-a_{n-1}\rvert \\ &\le p\cdot q^{n-2}\lvert a_2-a_1\rvert }$

comment.
Am I on the right track? Should I "get rid" of $p$? (and how)

Answer 1

You certainly are on the right track. Here's a hint:\begin{align}|a_8-a_5|&=|a_8-a_7+a_7-a_6+a_6-a_5|\\&\leqslant|a_8-a_7|+|a_7-a_6|+|a_6-a_5|\\&<(q^2+q+1)|a_6-a_5|\\&<(q^6+q^5+q^4)|a_2-a_1|\end{align}

Answer 2

Yes, you need to get rid of $p$ as $p$ can grow without bound. For $n>m,$ try to show

$$|a_n - a_m| \le (q^{n-2}+ q^{n-3}+ \cdots + q^{m-1})|a_2-a_1|.$$

That sum can be made small by choosing $m$ large enough, no matter how big $n$ is.