TITLE: Why does this process, when iterated, tend towards a certain number? (the golden ratio?)
QUESTION [13 upvotes]: 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.

REPLY [2 votes]: 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 $\displaystyle 1/L = \frac{1}{1+1/L}$).