Prove that $ \left(a_{n}\right)_{n=1}^{\infty} $ converges when $|a_{n+1}-a_{n}|<q|a_{n}-a_{n-1}|$ for $ 0<q<1 $

Question

I'm stuck on a homework question, and could really use some help. Here is said question:

"Assume that for every $n$ the following occurs: $|a_{n+1}-a_{n}|<q|a_{n}-a_{n-1}|$ when $ 0<q<1 $

Prove that the series $ \left(a_{n}\right)_{n=1}^{\infty} $ converges. Hint: use Cauchy sequence"

OK I'm really stuck on this one, I don't even have a clue where to begin.

Any help is appreciated. Thank you!

Answer 1

Hint Try to understand and complete the proof that $(a_n)$ is a Cauchy sequence.

For $n\ge m$ we have $$|a_n-a_m|=\left|\sum_{k=m}^{n-1} a_{k+1}-a_k\right|\le \sum_{k=m}^{n-1}| a_{k+1}-a_k|\le|a_1-a_0|\sum_{k=m}^{n-1}q^k\le|a_1-a_0|\frac{q^m}{1-q}\to0$$

Answer 2

Since $$ d_n:=|a_{n+1}-a_n|<q|a_n-a_{n-1}|=qd_{n-1} \quad \forall n \ge 2, $$ it follows that $$ d_n\le q^{n-2}d_2=q^{n-2}|a_2-a_1| \quad \forall n\ge 2 $$ For every $m, n \in \mathbb{N}$, say with $n>m$, we have \begin{eqnarray} |a_n-a_m|&\le&|a_n-a_{n-1}|+|a_{n-1}-a_{n-2}|+\ldots+|a_{m+1}-a_m|\\ &=&d_{n-1}+d_{n-2}+\ldots+d_m\\ &\le&(q^{n-3}+q^{n-4}+\ldots+q^{m-2})d_2\\ &=&\frac{q^m-q^n}{q^2(1-q)}d_2. \end{eqnarray} Thus $$ \lim_{m,n\to\infty}|a_n-a_m|=0, $$ i.e. $(a_n)_{n\ge 1}$ is a Cauchy sequence in $\mathbb{R}$, and therefore it is convergent.