Is $(a_n)_n$ with $|a_{n+2}-a_{n+1}| ≤ q|a_{n+1}-a_n|$ a Cauchy sequence?

Question

Let $0 < q < 1$ and $(a_n)_n$ be a squence with $|a_{n+2}-a_{n+1}| ≤ q|a_{n+1}-a_n|$ for all $n ∈ ℕ$.
I need to show that this is a Cauchy sequence. I'm not sure how to start this one, as we just had cauchy sequences in lecture. Can I maybe prove this by using induction?
How do start on a problem like this and where do I need to get to?

Answer 1

The increments $|a_{n+1}-a_n|$ are dominated by a geometric progression and therefore have a finite sum. Hence the sequence converges.

Answer 2

In order to prove that it is a Cauchy sequence you have to prove that $$|a_{n+p}-a_n|\longrightarrow 0,\text{ when }n\to\infty,p\to\infty,n,p\in\mathbb N$$

In your case , I 'll give you a HINT:

$$|a_{n+p}-a_n|=|a_{n+p}-a_{n+p-1}+a_{n+p-1}-\dots-a_{n+1}+a_{n+1}-a_n|\leq$$ $$|a_{n+p}-a_{n+p-1}|+\dots +|a_{n+1}-a_n|$$

Can you now continue? Express each difference above in terms of $q,a_0,a_1$

Answer 3

Hint:

• $|a_{n+1}-a_n|\leq q|a_{n}-a_{n-1}|\leq q^2|a_{n-1}-a_{n-2}|\leq\ldots\leq q^{n-1}\dfrac{1}{1-q}|a_{2}-a_{1}|$

• $|a_{n+2}-a_n|\leq |a_{n+2}-a_{n+1}|+ |a_{n+1}-a_{n}|\leq (q+1)|a_{n+1}-a_{n}|\\ \leq q(q+1)|a_{n-1}-a_{n-2}|\leq\ldots\leq q^{n-2}(q+1)|a_{2}-a_{1}|\leq q^{n-2}\dfrac{1}{1-q}|a_{2}-a_{1}|$

• $|a_{n+3}-a_n|\leq |a_{n+3}-a_{n+2}|+|a_{n+2}-a_{n+1}|+ |a_{n+1}-a_{n}|\leq\\ (q^2+q+1)|a_{n+1}-a_{n}|\leq q(q^2+q+1)|a_{n-1}-a_{n-2}|\leq\ldots\leq \\ q^{n-2}(q^2+q+1)|a_{2}-a_{1}|\leq q^{n-2}\dfrac{1}{1-q}|a_{2}-a_{1}|$

Now generalize that to show $$|a_{m}-a_n|\leq\ldots\leq q^{n-2}(q^{m-n}+\cdots+q^2+q+1)|a_{2}-a_{1}| \leq \\ q^{n-2}\dfrac{1}{1-q}|a_{2}-a_{1}|$$ and therefore $(a_n)$ is Cauchy (because $\displaystyle\lim_{n\to\infty}q^n=0$).