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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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marked as duplicate by Erick Wong, MJD, Asaf Karagila, Lukas Geyer, Micah Nov 21 '12 at 19:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

ive got up till the inductive step where $3 \leq 4-2/a_k < 4$ but dont know where to go from here? – jill Nov 21 '12 at 18:43
The same question, and more, has been asked and answered today. – André Nicolas Nov 21 '12 at 18:47
@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 $n \rightarrow \infty$ " – jill Nov 21 '12 at 19:49
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! – André Nicolas Nov 21 '12 at 19:56
Thank you soo much :D! – jill Nov 21 '12 at 20:13

2 Answers 2

Your induction hypothesis is that $3\le a_k<4$, and you want to use that to show that $3\le a_{k+1}<4$. This is just a matter of checking some simple algebra of inequalities: you know that $$a_{k+1}=4-\frac2{a_k}\;,$$ and you’re assuming that $3\le a_k<4$. How can you use that assumption to get information about $a_{k+1}$?

Start by seeing what it says about $\dfrac2{a_k}$: if $3\le a_k<4$, then

$$\frac13\ge\frac1{a_k}>\frac14\;,$$ so $$\frac23\ge\frac2{a_k}>\frac24=\frac12\;.$$

Now what does that say about $a_{k+1}=4-\dfrac2{a_k}$? Once again the inequalities have to be inverted, and we get


I’ll leave the last little bit of cleaning up to you.

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At the induction step, if $3 \leq a_k < 4$, then $$\dfrac14 < \dfrac1{a_k} \leq \dfrac13 \implies \dfrac12 < \dfrac2{a_k} \leq \dfrac23 \implies -\dfrac23 \leq -\dfrac2{a_k} < -\dfrac12 \implies 4-\dfrac23 \leq 4-\dfrac2{a_k} < 4-\dfrac12$$ This gives us that $$3 \dfrac13 < a_{k+1} < 3 \dfrac12$$

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thankyou :D!!!!!! – jill Nov 21 '12 at 18:50

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