Suppose that ${a_{n}}$ is a sequence such that, for all $n \in N$, $|a_n| < 2$, and $|a_{n+2} - a_{n+1}| \leq \frac{1}{8}|a_{n+1}^2 - a_{n}^2|$ . Prove that ${a_n}$ is a Cauchy sequence.
My thought is:
First. I know that I will factorize this term inside the modulus by the difference of squares rule $\frac{1}{8}|a_{n+1}^2 - a_{n}^2|$ , but I can not understand how I will use the following piece of information $|a_{n}| < 2$ that is given above.
Second. I feel like I will use this problem
let $0<r<1, M>0 $ and suppose that {a_n}is a sequence such that, for all $n \in N$ $|a_{n+1} - a_{n}| \leq Mr^n$ then the sequence is Cauchy. am I correct? But I am confused about the details.
Hint:
$$\frac{|a_{n+1}^2-a_n^2|}{8}=\left(\frac{|a_{n+1}-a_n|}{2}\right)\cdot\left(\frac{|a_{n+1}+a_n|}{4}\right).$$
Since $|a_n|<2$ for all $n$, what do you know about the second term of this product?