TITLE: On a recursive sequence exercise.
QUESTION [7 upvotes]: I have the following recursive sequence of which I want to prove the convergence:
$$x_{n+1} = \frac{x_n +1}{x_n +2 }$$ and $x_1 = 0$
I have proved that it is bounded above by $1$ and that it is increasing by taking the derivative, but I am told to do it without using derivatives.
How could I show that it's bounded and increasing using only elementary methods?
In particular I would like to prove that it is bounded by $\frac{\sqrt{5} - 1}{2}$ (this is the limit), but any additional ways to solve it are obviously very appreciated.

REPLY [2 votes]: Functions of the form $f:\; x \mapsto \dfrac{ax+b}{cx+d}$ are fractional linear transformations, and their iterates can be computed in closed form: 
$f^n: x \mapsto \dfrac{a_n x + b_n}{c_n x + d_n}$ where the matrix
$\pmatrix{a_n & b_n\cr c_n & d_n} = \pmatrix{a & b\cr c & d\cr}^n$.
In your case $$\pmatrix{a & b\cr c & d\cr} = \pmatrix{1 & 1\cr 1 & 2\cr}$$
and it's not hard to show that $$\pmatrix{a_n & b_n\cr c_n & d_n\cr} = \pmatrix{F_{2n-1} & F_{2n}\cr F_{2n} & F_{2n+1}}$$ where $F_n$ are the Fibonacci numbers.  In particular, with $\phi = \dfrac{1+\sqrt{5}}{2}$ is the "golden ratio", we have $F_n \sim \phi^n$ as $n \to \infty$.  With 
 $x_1 = 1$ we get $$x_n = f^{n-1}(1) = \dfrac{F_{2n-3} + F_{2n-2}}{F_{2n-2} + F_{2n-1}} = \dfrac{F_{2n-1}}{F_{2n}} \to 1/\phi = \dfrac{-1+\sqrt{5}}{2}$$