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We define a sequence of rational numbers {$a_n$} by putting $a_1=3$ and $a_{n+1}=4-\frac{2}{a_n}$ for all natural numbers. Put $\alpha = 2 + \sqrt{2}$

Could someone help me with a proof for the following.

Prove by induction on $n$ that $3 \leq a_n < 4$, where $a_1 = 3$ and $a_{n+1} = 4- 2/a_n$.

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    ive got up till the inductive step where $3 \leq 4-2/a_k < 4$ but dont know where to go from here?2012-11-21
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    The same question, and more, has been [asked and answered today.](http://math.stackexchange.com/questions/242014/we-define-a-sequence-of-rational-numbers-a-n-by-putting-a-1-3-and-a-n1)2012-11-21
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    @AndréNicolas ... i was just wondering because you answered the question on the other paper listed on your comment, how would you go about answering part e? "Deduce that $a_n \rightarrow \alpha$ as $n \rightarrow \infty$ "2012-11-21
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    It is immediate. By (d) we have $|a_n-\alpha|\le\frac{|a_1-\alpha|}{4^{n-1}}$. But since $4^{n-1}$ gets huge for large $n$, $\frac{|a_1-\alpha|}{4^{n-1}}\to 0$, so $|a_n-\alpha|\to 0$, and therefore $a_n\to\alpha$ as $n\to\infty$. The author of the problem has broken it up into bite-sized pieces!2012-11-21

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