Real analysis: Sequences question

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jinbaw
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Homework Statement


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


Homework Equations





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 [tex]X^2_{n+1} - X^2_n[/tex].

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

Iterating, I reached [tex]|X_{n+2} - X_ {n+1}| \leq \frac{|X_2 - X_1|}{2^n}[/tex]
I'm not sure if I continued in the right track.. but I'm stuck.
Thanks for your input.
 
Now go back to trying to show the sequence is Cauchy by adding and subtracting terms in |xn-xm|. Use the triangle inequality.
 
Or could even look at the sequence of partial sums, getting it bounded above.