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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.

Intuition
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1 Answers1

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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?

Carl Schildkraut
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  • less than 4 >>>> right? after using the triangle inequality – Intuition Sep 17 '18 at 02:52
  • I am not sure how can I apply the problem I mentioned above in second ..... could you give me more details please? how to determine the $N$? – Intuition Sep 17 '18 at 03:02
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    @hopefully From this you can get $|a_{n+2}-a_{n+1}| < \frac{1}{2}\ |a_{n+1}-a_n|.$ This can be used inductively to apply your second idea. – Carl Schildkraut Sep 17 '18 at 03:21