Why does this process, when iterated, tend towards a certain number? (the golden ratio?) 
*

*Take any number $x$ (edit: x should be positive, heh)

*Add 1 to it $x+1$

*Find its reciprocal $1/(x+1)$

*Repeat from 2


So, taking $x = 1$ to start:


*

*1

*2 (the + 1)

*0.5 (the reciprocal)

*1.5 (the + 1)

*0.666... (the reciprocal)

*1.666... (the + 1)

*0.6 (the reciprocal)

*1.6 (the + 1)

*0.625

*1.625

*0.61584...

*1.61584...

*0.619047...

*1.619047...

*0.617647058823..


etc.
If we look at just the "step 3"'s (the reciprocals), we get:


*

*1

*0.5

*0.666...

*0.6

*0.625

*0.61584...

*0.619047...

*0.617647058823..


This appears to always converge to 0.61803399... no matter where you start from.  I looked up this number and it is often called "The golden ratio" - 1, or $\frac{1+\sqrt{5}}{2}-1$.


*

*Is there any "mathematical" way to represent the above procedure (or the terms of the second series, of "only reciprocals") as a limit or series?

*Why does this converge to what it does for every starting point $x$?



edit: darn, I just realized that the golden ratio is actually 1.618... and not 0.618...; I edited my answer to change what the result is apparently (golden ratio - 1).
However, I think I could easily make it the golden ratio by taking the +1 "steps" of the original series, instead of the reciprocation steps of the original series:


*

*2

*1.5

*1.666...

*1.6

*1.625

*1.61584...

*1.619047...

*1.617647058823..


which does converge to $\frac{1+\sqrt{5}}{2}-1$
Explaining either of these series is adequate as I believe that explaining one also explains the other.
 A: This was supposed to be an addendum to Hans's comment, but it got too long.
The iteration Justin considered formally generates a continued fraction:
$$\cfrac{1}{1+\cfrac{1}{1+\dots}}$$
and as I mentioned in this answer, the numerators and denominators of the nth convergent of a continued fraction can be computed recursively.
If we apply the formula in that answer to this situation, we get
$$\begin{bmatrix}C_n\\\\D_n\end{bmatrix}=\begin{bmatrix}C_{n-1}\\\\D_{n-1}\end{bmatrix}+\begin{bmatrix}C_{n-2}\\\\D_{n-2}\end{bmatrix}$$
with initial conditions
$$\begin{bmatrix}C_{-1}\\\\D_{-1}\end{bmatrix}=\begin{bmatrix}1\\\\0\end{bmatrix},\qquad \begin{bmatrix}C_{0}\\\\D_{0}\end{bmatrix}=\begin{bmatrix}0\\\\1\end{bmatrix}$$
From here, the setup is now similar to Robin's answer, since the two recursions in fact generate the Fibonacci numbers: $C_n=F_n$ and $D_n=F_{n+1}$
Your continued fraction then is the limit
$$\lim_{n\to\infty}\frac{F_n}{F_{n+1}}$$
and the equivalence to Robin's answer is due to the Binet formula:
$$F_n=\frac{\varphi^n-\left(-\varphi^{-1}\right)^n}{\sqrt{5}}$$
(which in fact can be derived from Robin's answer).
Substituting that into the limit and evaluating gets you the answer.
A: You are iterating the operation
$$x\mapsto\frac1{x+1}.$$
We can represent this in matrix terms. Set
$$A=\begin{pmatrix}
0&1\\
1&1\end{pmatrix}.$$
If
$$A
\begin{pmatrix}
x\\
1\end{pmatrix}
=\begin{pmatrix}
y_1\\
z_1\end{pmatrix}$$
then $1/(x+1)=y_1/z_1$. After $n$ iterations one gets
$$A^n
\begin{pmatrix}
x\\
1\end{pmatrix}
=\begin{pmatrix}
y_n\\
z_n\end{pmatrix}$$
and $x_n=y_n/z_n$ is got by applying the map $x\mapsto 1/(x+1)$ $n$ times to $x$.
Naturally you won't be surprised to find that the eigenvalues of
$A$ are $\tau=\frac12(1+\sqrt5)$ and $\tau=\frac12(1-\sqrt5)$.
The eigenvectors are $v=(1\ \tau)^t$ and $w=(1\ \tau')^t$. We can write
$$\begin{pmatrix}
x\\
1\end{pmatrix}=av+bw$$
and so
$$\begin{pmatrix}
y_n\\
z_n\end{pmatrix}=a\tau^n v+b\tau'^nw.$$
Now $\tau>1>|\tau'|$
so that for large $n$, $y_n$ and $z_n$ are very close to $a\tau^n$and $a\tau^{n+1}$
so that $x_n\to\infty$ if $a=0$. The only exception is when $a=0$
which only arises for $x=-\frac12(1+\sqrt5)$. (I bet you didn't test that one!)
A: Here is another way of looking at it.
First consider the special case of starting with $1$.
Consider what happens when  $\displaystyle x = \frac{f_n}{f_{n+1}}$ where $f_n$ is the $n^{th}$ fibonacci number.
You get $$\frac{1}{\frac{f_n}{f_{n+1}} + 1} = \frac{f_{n+1}}{f_n + f_{n+1}} = \frac{f_{n+1}}{f_{n+2}}$$
Since $\displaystyle 1 = \frac{f_1}{f_2}$
we see that after $n$ iterations, $\displaystyle x = \frac{f_{n+1}}{f_{n+2}}$
This can be generalized to any other starting value, by using a Fibonacci like sequence, which satisfies the recurrence $\displaystyle a_{n+2} = a_{n+1} + a_{n}$ and choosing appropriate $a_{2}$ and $a_{1}$ so that $\displaystyle \frac{a_1}{a_2}$ is the initial guess for $x$. 
The $n^{th}$ value for $x$ will be given by $\displaystyle \frac{a_n}{a_{n+1}}$
The general formula for such sequences is given by $a_{n} = A\alpha^n + B\beta^n$ where $\alpha,\beta$ are roots of $\displaystyle z^2 = z + 1$ and thus the limit of $\displaystyle \frac{a_n}{a_{n+1}}$ can be easily found, which will be one of $1/\alpha$ or $1/\beta$ (which you can also see, by assuming there is a limit $1/L$ and setting $\displaystyle 1/L = \frac{1}{1+1/L}$).
A: We want to show that the function $f(x) = \frac{1}{1+x}$ has a unique fixed point to which it converges when iterated on positive $x$.  If $x$ is positive, then $f(x)$ is between $1$ and $0$, so $f(f(x))$ is between $1$ and $\frac{1}{2}$.  It is not hard to see that in fact $f$ fixes the interval $\left[ \frac{1}{2}, 1 \right]$.  Now, for $x, y$ in this interval,
$$\left| \frac{1}{x+1} - \frac{1}{y+1} \right| = \left| \frac{y-x}{(1+x)(1+y)} \right| \le \frac{4}{9} |x - y|$$
so on this interval $f$ satisfies the conditions of the Banach fixed point theorem.  The unique fixed point to which everything always converges is the unique solution to $f(x) = x$, which you have already found.
A: Define the function $f(x) = \frac1{1+x} $ so the process you defined is just going from x to f(x).
let $x_0\in\mathbb{R}$ and $x_{i+1} = f(x_i)$ then your claim is that $\lim\; x_i = \frac {1+\sqrt{5}}{2} $.
if this series converges then
$x = \lim\; x_i = \lim\; f(x_{i-1}) = f( \lim\; x_{i-1} ) = f(x)$ so the limit is $x=\frac{1}{1+x}$ or in other words you have $x^2+x-1=0$. the solutions are $\frac{-1\pm \sqrt{5}}{2}$ 
this isn't exactly the golden ratio. if you take $\frac1{x-1}$ instead then you'll have it. you still need to find out for which $x_0$ this sequence converges - this could be done using numerical analysis (sorry, I'm a bit rusty on this, so I'll leave this part to someone else).
