How to prove rational sequence converges to irrational number

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potatocake
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Homework Statement
Construct (with a proof) a rational sequence xn that converges 1+sqrt(2)/2
Relevant Equations
x0 = 5/4
f(x) = 1 + 1/4x
xn = f(xn-1) for n = 1,2,....
I attempted to solve it

$$ x = \frac {1}{4x} + 1 $$
$$⇒ x^2 -x -\frac{1}{4} = 0 $$
$$⇒ x = \frac{1±\sqrt2}{2} $$

However, I don't know the next step for the proof.
Do I need a closed-form of xn+1or do I just need to set the limit of xn and use inequality to solve it?

If I have to use the inequality sign, how can I set the interval of xn?
 

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Given what you've written down, you should be able to conclude that if ##x_n## converges, it converges to
$$ \frac{1\pm \sqrt{2}}{2}.$$

This leaves some open questions
1.) Does your sequence consist of rational numbers?
2.) Does your sequence converge?
3.) Which root does it converge to?
 
It is obvious that if [itex]x_n > 0[/itex] then [itex]x_{n+1} > 1 > 0[/itex].

You can also show directly that [tex]\left| x_{n+1} - \frac{1 + \sqrt{2}}{2}\right|<br /> = \left|\frac{1 - \sqrt{2}}{2x_n}\right|\left|x_n - \frac{1 + \sqrt{2}}{2}\right|[/tex] and hence that [itex]|x_n - (1 + \sqrt{2})/2|[/itex] is strictly decreasing for appropriate choices of [itex]x_0[/itex].
 
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