# 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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## Answers and Replies

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.
Thanks for your input.

Dick
Science Advisor
Homework Helper
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.