solution of recurrence relation $x_{n+1} = \frac {1}{x_n + 1}$ $$x_{n+1} = \frac {1}{x_n + 1}; x_1 > 0$$
How to transform it into the form $x_n = $? I need the solution in order to check if it converges at any $x_1 > 0$.
 A: hint 
One nice way to do linear fractional recurrences is to use matrices.  If $a/b=x$, then $c/d = 1/(x+1)$, where
$$
\left(\begin{aligned}0\qquad 1 \cr 1\qquad 1\end{aligned} \right)
\left(\begin{aligned}a\cr b\end{aligned}\right) = 
\left(\begin{aligned}c\cr d\end{aligned}\right)
$$
So we can get a formula for $x_n$ if we know a formula for the $n$th power of that $2 \times 2$ matrix.
A: Some hints:


*

*First of all you can prove by induction that all of the $x_n$'s are strictly positive.

*Using that, prove that they are also lesser than 1.

*From the above two steps, the sequence $\{ x_n \}$ lies in the compact interval $[0,1]$,so it has a convergent subsequence $\{ x_{n_k} \}$

*Prove that $\lim_{n \to \infty} x_n$ exists and is equal to $\lim_{k \to \infty} x_{n_k}$. 

A: Try
$$a(n)\to \frac{\mathcal C \;\left(\frac{1}{2}
   \left(1+\sqrt{5}\right)\right)^n+\left(\frac{1}{2}
   \left(1-\sqrt{5}\right)\right)^n}{\mathcal C \; \left(\frac{1}{2}
   \left(1+\sqrt{5}\right)\right)^{n+1}
   +\left(\frac{1}{2}
   \left(1-\sqrt{5}\right)\right)^{n+1}}$$
A: Let $a=\lim_{n\to\infty}x_n$. Then
\begin{align}
&a=\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}\\&a=\frac{1}{1+a}\\&a(a+1)=1\\&a^2+a-1=0\\&a=\frac{-1\pm\sqrt{5}}{2}
\end{align}
Since there is no negative term or subtraction in the sequence, we omit the negative solution.
$\displaystyle \therefore a=\lim_{n\to\infty}x_n=\frac{\sqrt{5}-1}{2}$
A: If you are interested in finding the limit, assume $\lim_{n\to \infty} a_n = a$, then
$$ a=\frac{1}{a+1} \implies a^2+a-1=0 \implies a=\frac{-1 \bar{+} \sqrt{5}} {2}. $$
Now, you should pick up the right limit. 
