Solution to $x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}}$ 
Possible Duplicate:
Why does this process, when iterated, tend towards a certain number? (the golden ratio?) 

Please post your favorit solution to the following 
Compute $x=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\ldots}}}$
Thank you
 A: Look at the equation $$x^2-x-1=0$$
It is clear is solution is not zero. Thus, write
$$x^2=x+1$$
This is equivalent to
$$x=1+\frac 1 x$$
Using this recursively
$$x=1+\frac 1 {1+\frac 1 x}$$
Thus, define $\{ x_n\} $ recursively as $x_0=1$ and $x_n=1+\dfrac{1}{x_{n-1}}$
Show the sequence is positive, increasing for $n>3$ and bounded, thus it converges to $\ell$. You can then show that $\lim x_n=\ell = x$, where $x$ is the positive solution of the first equation discussed, namely
$$\phi = \frac{\sqrt 5 +1}{2}$$
A: Denote by $x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}}$. Because the fraction goes on forever, the denominator of the second term in the RHS is identical to $x$. Thus, we have $x=1+1/x$ or the same as $x^2=1+x$. Thus, $x=\frac{\sqrt 5-1}{2}$.
A: Here is another, more pedestrian, and sketchy, way of doing this:
Let the sequence $\frac{p_n}{q_n}$ be defined as follows:
$$p_1=q_1=1$$
$$\frac{p_n}{q_n}=1+\frac 1{\frac{p_{n-1}}{q_{n-1}}} = \frac{p_{n-1}+q_{n-1}}{p_{n-1}}$$
It is clear that this computes terminated versions of the continued fraction and we have $$q_n=p_{n-1} \text{  and  } p_n=p_{n-1}+q_{n-1}=p_{n-1}+p_{n-2}$$
It follows that numerator and denominator are successive Fibonacci numbers, and the limit ratio is the golden ratio.
There is an efficient discussion of continued fractions in Hardy & Wright "An Introduction to the Theory of Numbers" which shows that successive estimates from terminated fractions lie either side of the limit and converge to it.
