Different ways of computing $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$ 
Possible Duplicate:
On the sequence $x_{n+1} = \sqrt{c+x_n}$  

I am wondering how many different solutions one can get to the following question:

Calculate $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$

Post your favorite solution please.
 A: Consider the sequence $$x_n = \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots \text{ n times}}}}$$ i.e. $$x_{n+1} = \sqrt{1+x_n}.$$
For instance, $x_1 = \sqrt{1}$, $x_2 = \sqrt{1+\sqrt{1}}$, $x_3 = \sqrt{1+\sqrt{1+\sqrt{1}}}$.
We will prove that the sequence is monotonically increasing and is bounded above by $\phi$ where $\phi = \dfrac{1+\sqrt{5}}{2}$ satisfies $\phi^2 = 1 + \phi$ and $\phi > 0$.
Boundedness
First we will prove that the sequence is bounded above by $\phi$. The proof follows from induction.
For $n=1$, $x_n = 1 = \dfrac{1+1}{2} < \dfrac{1+\sqrt{5}}{2} = \phi$. Hence, the claim is true for $n=1$.
Now assume it is true for some $k \in \mathbb{N}$.
$x_k < \phi \implies 1+x_k < 1 + \phi = \phi^2$. (Since $\phi = \dfrac{1+\sqrt{5}}{2}$ satisfies $\phi^2 = 1 + \phi$).
Hence, $x_{k+1} = \sqrt{1+x_k} < \phi$ and thereby the sequence is bounded above.
By a similar argument, we also have that $x_n$ is bounded below by $1$. Hence, we now have that $$1 \leq x_n < \phi.$$
Monotonically increasing
We will now prove that the sequence is monotonically increasing i.e. $x_n < x_{n+1}$.
Consider $x_{n+1}^2 - x_n^2$. We have $x_{n+1}^2 - x_n^2 = (1+x_n) - x_n^2 = \dfrac54 - \left( x_n - \dfrac12\right)^2$.
The quadratic equation $f(x) = 1 + x - x^2$ is positive whenever $x \in \left( \dfrac{1-\sqrt{5}}{2}, \dfrac{1+\sqrt{5}}{2}\right)$.
Since we proved earlier that $x_n \in \left[1,\dfrac{1+\sqrt{5}}{2}\right)$, we have that $x_{n+1}^2 - x_n^2 = 1 + x_n - x_n^2 > 0$.
Since $x_n$'s are positive, we have that $$x_{n+1} > x_n.$$
Hence, now combining both we get that $$1 \leq x_n < x_{n+1} < \phi$$
Hence, by monotone sequence theorem, (or equivalently by completeness of $\mathbb{R}$), the sequence $\{x_n\}_{n=1}^{\infty}$ converges.
Once we know that the sequence converges, we can then make use of the following limit rules.


*

*If $y_n \rightarrow L$, then $y_{n+1} \rightarrow L$.

*If $y_n \rightarrow L$, then $1+y_n \rightarrow 1+L$.

*If $z_n \rightarrow L$, and $z_n > 0$, then $L \geq 0$.

*If $z_n \rightarrow L$, and $z_n > 0$, then $\sqrt{z_n} \rightarrow \sqrt{L}$.


Let us now assume that $x_n \rightarrow L$, then making use of the above limit rules, we get that $$L = \sqrt{1+L}$$ Since $L \geq 0$ (because $x_n > 0)$, solving the quadratic we get that $$L = \dfrac{1+\sqrt{5}}2$$
A: Assuming convergence, here is one:
$$S=\sqrt{1+\sqrt{1+...}}=\sqrt{1+S}\implies S^2=1+S \implies S=\frac{1+\sqrt{5}}{2}$$
A: If
$$\phi = \sqrt{1+\sqrt{1+...}}$$
then
$$\phi^2= 1+\sqrt{1+\sqrt{1+...}}=1+\phi$$
hence
$$\phi^2-\phi-1=0$$
$$\phi=\frac{1\pm\sqrt{5}}{2}$$
EDIT:
the sequence $\phi_n=\sqrt{1+\phi_{n-1}}$ is apparently increasing. $\phi_1<2$. Assume $\phi_{k-1}<2$, $k>2$. Then $1+\phi_{k-1}<3\implies \phi_k=\sqrt{1+\phi_{k-1}}<\sqrt{3}<2$, hence the sequence is bounded.
