$\sqrt{1 + \sqrt{1 + \sqrt{1 + ...}}} = \frac{1+\sqrt{5}}{2} = \phi$, is this a coincidence? I was playing around with square roots today when I "discovered" this.  
$\sqrt{1 + \sqrt{1 + \sqrt{1 + ...}}} = x$
$\sqrt{1 + x} = x$
$1 + x = x^2$ 
Which, via the quadratic formula, leads me to the golden ratio.  
Is there any significance to this or is it just a random coincidence?  
 A: Any expression of the form $\sqrt{n+\sqrt{n+\sqrt{n+\ldots}}}$ will have a valueof the form 
$$
\frac{-1\pm\sqrt{1+4n}}{2}
$$
so it is no coincidence that you get a simple answer like you got.
As to whether it is a coincidence that this is the golden ratio, pretty much the defining expression for the golden ratio comes from a rectangle out of which a square is cut, leaving a similar rectangle -- and that diagram immediately gives you 
$$ \frac{1+x}{x} = \frac{x}{1} \implies 1+x = x^2$$
So although a bit of opinion creeps in, I'd say this is no coincidence.
A: If you play around a little more, you will also notice that:
$$
\frac{1+\sqrt{5}}{2} = 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+ \ldots} } } } 
$$
Which simplifies to $x = 1+ \frac 1x \implies x^2=x+1$.
It's no coincidence. I mean to say, it comes directly from the equation itself. 
Just to give you another example: The equation $x^2 = 4+x$ is satisfied by the fraction $\frac{1+\sqrt{17}}{2}$. Now, we can use the same logic to extend this fellow:
$$
x = \sqrt{4 + x} = \sqrt{4 + \sqrt{4 + \sqrt{4 + \sqrt{4 + \ldots}}}}
$$
While at the same time, this also expands as a continuous fraction, namely:
$$
x = 1 + \frac{4}{x} = 1 + \frac{4}{1 + \frac{4}{1 + \frac{4}{1 + \frac{4}{1 + \ldots} } } } 
$$
You see, it's not a coincidence, yet it's wonderful.
The question arises: Can we do this with other quadratic polynomials?
Take for example, $ax^2+bx+c=0$. Then $ax^2 = -bx-c$ and $x^2 = -\frac{b}{a}x -\frac{c}{a}$.
This will expand now in an interesting way:
$$
x = \sqrt{-\frac{c}{a}-\frac{b}{a}x} = \sqrt{-\frac{c}{a}-\frac{b}{a}\sqrt{-\frac{c}{a}-\frac{b}{a}\sqrt{-\frac{c}{a}-\frac{b}{a} \ldots}}}
$$
And as a continuous fraction:
$$
x = -\frac{b}{a} - \frac{c}{ax} =-\frac{b}{a} - \frac{c}{a(-\frac{b}{a} - \frac{c}{a(-\frac{b}{a} - \frac{c}{a \ldots} )}) }  
$$
That is your license to play around. Please do so. Also, see what you get if $ax^3+bx^2+cx+d=0$, and if you can find something interesting here do comment.
A: Let $a_n$ include $n$'s 1. Then $a_{n+1}=\sqrt{1+a_n}$. One can show by induction that $a_n$ is increasing and is bounded from above by $3$. Induction steps:
$$
a_{n+1}=\sqrt{1+a_n}\geq\sqrt{1+a_{n-1}}=a_n; a_{n+1}=\sqrt{1+a_n}<\sqrt{1+3}=2<3.
$$
So there exists $L=\lim a_n$ satisfying $L=\sqrt{1+L}$, which implies $L$ equals the value you find.
