TITLE: Convergence of $a_{n+1}=\frac{1}{1+a_n}$
QUESTION [9 upvotes]: We define
$$a_{n+1}=\frac{1}{1+a_n}, a_0=c> 0$$ 
I stumbled across that sequence and Mathematica gives me that it converges against $\frac{1}{2} \left(\sqrt{5}-1\right)$ which doesn't even depend on $a_0$. Normally I show that sequences converge by seeing that they are monotonic and then the limit can be easily found by setting $a_n=a_{n+1}$, but this one seems to be alternating. Also by checking OEIS I noticed that $a_n=1/g(n)$ where $g(n)$ gives the $n+1$'th Golden Rectangle Number.
I would be glad if someone can show me how to show that such sequences converge and how to find the limit, also maybe this is a well known sequence, as it has a quite simple form, too bad that google is very bad for looking for sequences and OEIS doesn't mention anything.

REPLY [9 votes]: For $x\geq 0$, $f(x)=1/(1+x)$ is a monotonically decreasing function, bounded between 0 and 1. The fixed point (if it exists) of the iterative map defined by your function, is the solution of $x=\frac{1}{1+x}$. There is only one solution for $x\geq 0$, and that is $\tilde{x}=(\sqrt{5}-1)/2$.
If we look at the evolution of $a_n=f(f(f(...(a_0))))$, we get 
$$a_n=\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+a_0}}}}}$$
and as $n\to\infty$, what $a_0$ was becomes irrelevant. In other words, no matter the value of $a_0$, it converges to a fixed point, which has to be the only fixed point, $\tilde{x}$. Hence $\tilde{x}$ is an attractive fixed point and also the limit of the sequence.

The Mathematica code that produced the above figure is below (it's modified from an answer to a question on SO)
Clear[fixedPoint]
fixedPoint[a0_, nSteps_] := 
 Module[{f, a, n}, 
  f = RecurrenceTable[{a[n] == 1/(1 + a[n - 1]), a[1] == a0}, 
    a, {n, 1, nSteps}];
  Plot[{1/(1 + x), x}, {x, 0, 1}, 
   PlotStyle -> {Darker@Blue, {Dashed, Black}}, 
   Epilog -> {Darker@Green, 
     Line[Riffle[Partition[f, 2, 1], {#, #} & /@ Rest[f]]]}]]