How does ($x_n$) converge for $x_1=1$, $x_{n+1}=\frac{1}{x_n+3}$ for $n=1,2,\ldots$? 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.
 A: Hint: For $x,y \geq 0$ we have $\left\vert\frac{1}{x+3} - \frac{1}{y+3}\right\vert = \left\vert\frac{y-x}{(x+3)(y+3)}\right\vert \leq \frac{1}{9}\vert x-y\vert$.
A: 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 $x_{2n+3}>x_{2n+1}$.
A: 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 $|a/b|&lt1$ find the limit.  
