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bonfire09

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## Homework Statement

If ##x_1 < x_2## are arbitrary real numbers and ##x_n=\frac{1}{2}(x_{n-2}+x_{n-1})## for## n > 2##, show that ##(x_n)## is convergent.

## Homework Equations

Definition of Contractive Sequence: We say that a sequence ##X=(x_n)## of real numbers is contractive if there exists a constant ##C##; ##0 < C < 1##, such that ##|x_{n+2}-x_{n+1}|\leq C|x_{n+1}-x_{n}|## for all ##n\in\mathbb{N}##. The number ##C## is called the constant of the contractive sequence.

## The Attempt at a Solution

I know this sequence is not monotone so I think I need to use contraction. All I have is that ## |x_{n+2}-x_{n+1}|=| \frac{1}{2}(x_{n}+x_{n+1}-\frac{1}{2}(x_{n-1}+x_n|=|\frac{1}{2}x_{n+1}-\frac{1}{2}x_{n-1}|## and ##|x_{n+1}-x_n|=|\frac{1}{2}(x_{n-1}+x_n)-\frac{1}{2}(x_{n-2}+x_{n-1}|=\frac{1}{2}|x_n-x_{n-2}|##. But I can't seem to show that ##\frac{1}{2}|x_{n+1}-x_n|\leq \frac{1}{2}|x_n-x_{n-2}|## which I think is not possible but I am not sure. I think cauchy might work too but not sure how to apply it here.

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