# Real analysis: Sequences question

## Homework Statement

If Xn is bounded by 2, and $$|X_{n+2} - X_{n+1}| \leq \frac{|X^2_{n+1} - X^2_n|}{8}$$, prove that Xn is a convergent sequence.

## The Attempt at a Solution

I believe the solution lies in proving Xn a Cauchy sequence, but I'm not sure how to work it out. I considered |Xn - Xm| adding and subtracting the terms between m and n but i got stuck.
I also tried to check for telescoping, with no luck.

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Factor $$X^2_{n+1} - X^2_n$$.

Then get an upper bound for $$\frac{|X^2_{n+1} - X^2_n|}{8}$$.

$$\frac{|X^2_{n+1} - X^2_n|}{8} = \frac{|(X_{n+1} + X_n)(X_{n+1} - X_n)|}{8} \leq \frac{|X_{n+1} - X_n|}{2}$$

Iterating, I reached $$|X_{n+2} - X_ {n+1}| \leq \frac{|X_2 - X_1|}{2^n}$$
I'm not sure if I continued in the right track.. but I'm stuck.