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Show that the sequence ($x_n$) defined by $$x_1=1\quad \text{and}\quad x_{n+1}=\frac{1}{x_n+3} \quad (n=1,2,\ldots)$$ converges and determine its limit ?

I try to show ($x_n$) is a Cauchy sequence or ($x_n$) is decreasing (or increasing) and bounded sequence but I fail every step of all.

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What do you know? What did you try? Where are you stuck? What similar (simpler) questions can you solve? –  Did Mar 25 '12 at 15:10
    
Hint: draw the graph of $f:x\mapsto1/(x+3)$ on $0\leqslant x\leqslant1$ and try to follow the sequence $(x_n)$ on the graph. –  Did Mar 25 '12 at 15:14
    
What happened to the determine its limit part of your question? Hmmm... –  Did Mar 25 '12 at 15:35
    
unfortunately I try induction to show ($x_n$) is decreasing (or increasing) but ($x_n$) I couldnt find, so I think this is not the right way and I couldnt find a suitable m value for $|x_n-x_m|$ for ($x_n$) is a Cauchy sequnce –  Shamill Mar 25 '12 at 15:38
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You can show (by induction e.g.) that $(x_{2n})$ is decreasing and that $(x_{2n+1})$ is increasing. You can also show that $(x_n)$ is bounded below by 0 and bounded above by 1. Then you can show that both the "even" and "odd" sequences converge to the same number (both limits satisfy $L={1\over L+3}$, $L\ge0$). –  David Mitra Mar 25 '12 at 15:43

3 Answers 3

up vote 4 down vote accepted

Hint: For $x,y \geq 0$ we have $\left\vert\frac{1}{x+3} - \frac{1}{y+3}\right\vert = \left\vert\frac{y-x}{(x+3)(y+3)}\right\vert \leq \frac{1}{9}\vert x-y\vert$.

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You can easily show that $(x_n)$ is bounded below by 0 and bounded above by 1.

You can then show (by induction e.g.) that $(x_{2n})$ is decreasing and that $(x_{2n+1})$ is increasing. Then you can argue that the sequence $(x_{2n})$ converges to some number $L$ and that the sequence $(x_{2n+1})$ converges to some number $M$.

Now, since $x_{n+1} ={1-x_{n+1} x_n\over3}$, it follows that $(x_n)$ converges, to $b$, say. Then from the recursion formula, we must have $b={1\over b+3}$; solving this equation we see that $b$ is its positive solution $b={-3\over2}+{\sqrt{13}\over2}$.


For the induction argument to show that $(x_{2n})$ is decreasing and $(x_{2n+1})$ is increasing:

Verify that $x_1<x_3$ and that $x_2>x_4$.

Assume that both $x_{2n-1}<x_{2n+1}$ and $x_{2n-2}>x_{2n}$ hold.

Then show that $x_{2n+2}<x_{2n}$. Using this result, show that $x_{2n+3}>x_{2n+1}$.

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You can also find the explicit form of $a_n.$ The following argument is taken from Kaczor-Novak, Problems in Mathematical Analysis, vol.1, AMS pub. p.228

The equation $x^2+3x-1=0$ has two solutions $a > 0 >b.$ It is easy to observe $$ \frac{x_{n+1}-a}{x_{n+1}-b}=\frac{a}{b} \frac{x_n-a}{x_n-b}. $$ Write separately numerator and the denominator and simplify. Proceeding inductively we come to $$ \frac{x_{n+1}-a}{x_{n+1}-b} = \left(\frac ab\right)^n \frac{x_1-a}{x_1-b}. $$ Solve in term of $x_{n+1}$ and using $|a/b|<1$ find the limit.

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