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Let $0<x_1<1$. If $$x_{n+1}=\frac{x_n+3}{3x_n+1},n\in N$$ (1) Show that $$x_{n+2}=\frac{5x_n+3}{3x_n+5}$$ (2) Hence show that $\lim ~ x_n$ exists [i.e from part (1)]

(3) Find $\lim ~ x_n$

Help me in showing the existence part .I can show the part (1) .

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Hint: show that $a_{2n}>a_{2n+2}$, $a_{2n-1}<a_{2n+1}$ and $a_{2n-1}<a_{2n}$ by induction and (1). –  tetori Jan 25 '13 at 18:11

2 Answers 2

up vote 2 down vote accepted

Hint: Show that $(x_{2n-1})_{n\geqslant1}$ is increasing (and bounded above by $1$) and $(x_{2n})_{n\geqslant1}$ is decreasing (and bounded below by $1$).

To do that, you might want first to show that the function $u:x\mapsto\frac{5x+3}{3x+5}$ is increasing with $u(x)\gt x$ if $x$ is in $(0,1)$ and $u(x)\lt x$ if $x$ is in $(1,+\infty)$.

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I understand your last paragraph. How I show the sequence is increasing/decreasing and bounded above/below $1$. –  A.D Jan 25 '13 at 18:31
    
For example, $x_1\lt1$ and $x\lt u(x)\lt1$ for every $x$ in $(0,1)$ hence $x_1\lt x_3\lt1$. Can you carry on? –  Did Jan 25 '13 at 20:09
    
How can I show $x_2 >1$? –  A.D Jan 26 '13 at 16:31
    
Well, the function $v:x\mapsto(x+3)/(3x+1)$ sends the interval $(0,1)$ to $___$. –  Did Jan 26 '13 at 17:04
    
Thanks. Now I fully understand. –  A.D Jan 26 '13 at 18:18

Let $f(x)$ be $f(x)=\frac{x+3}{3x+1}$. Then for $n\geq2$, $$(\underbrace{f\circ f\cdots \circ f}_{(n-1) times})(x_1)=x_n$$ And let A a $2\times 2$ matrix such that $$A=\begin{pmatrix} 1 & 3\\ 3 & 1 \end{pmatrix}$$. By Cayley–Hamilton theorem, $A^2-2A-8E=0$, and let $X$ a $2\times 2$ matrix. And make an identical equation. (a,b are real numbers and $n\geq2$) $$X^{n-1}=(X^2-2X-8E)Q(X)+aX+bE$$ Substitute $4E$ and $-2E$, we get 2 simultaneous equations. $$4a+b=4^{n-1}$$ $$-2a+b=(-2)^{n-1}$$ Therefore, $a=\frac{4^{n-1}-(-2)^{n-1}}{6}$ and $b=\frac{4^{n-1}-(-2)^n}{3}$. Substitute A into the identical equation, $$A^{n-1}=aA+bE=-(-2)^{n-2}\begin{pmatrix} (-2)^{n-1}+1 & (-2)^{n-1}-1 \\ (-2)^{n-1}-1 & (-2)^{n-1}+1 \end{pmatrix}$$ For $n\geq2$, $$x_n=\frac{((-2)^{n-1}+1)x_1+(-2)^{n-1}-1}{((-2)^{n-1}-1)x_1+(-2)^{n-1}+1}$$ Therefore, we can know $$\lim_{n\rightarrow \infty}x_n=1$$. Sorry for my poor English and typing skill.

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