When does $\sqrt{x+\sqrt{x+1+\sqrt{x+2+...}}}=0$? Consider the function $f$ defined as the limit of the functions
$$f_0(x)=\sqrt{x}$$
$$f_1(x)=\sqrt{x+\sqrt{x+1}}$$
$$f_2(x)=\sqrt{x+\sqrt{x+1+\sqrt{x+2}}}$$
$$...$$
so that $f(x)$ is defined iff $f_n(x)$ is defined for some $n$. The unique root $x_0$ of the function $f$ satisfies $f(x_0)=0$, and it can be alternatively expressed as the limit of the roots of $f_0, f_1, f_2, ...$. See the graphs below:

Can anyone find an expression equal to this limit? I realize that the chances of something nice and closed-form are slim - can we find a series, integral, or even nested radical representation of the real root of $f(x)$?

Not sure if this is helpful, but it may be fruitful to consider a sequence of functions $g_n(x)$ where $g_0(x) = x$ and
$$g_n(x) = [g_{n-1}(x)]^2 - x - n$$
In particular, 
$$\lim_{n \to \infty} g_n(x_0) = 0.$$
I found this by setting
$$f_n(x) = 0$$
and moving all roots to the other side. For example, with $n = 2$,
$$\begin{align*}
0 &= \sqrt{x + \sqrt{x + 1 + \sqrt{x + 2 + \sqrt{x+3}}}}, \\
-x &= \sqrt{x + 1 + \sqrt{x + 2 + \sqrt{x+3}}}, \\
x^2 - x - 1 &= \sqrt{x+2 + \sqrt{x+3}}, \\
(x^2 - x - 1)^2 - x - 2 &= \sqrt{x+3}, \\
((x^2 - x - 1)^2 - x - 2)^2 - x - 3 &= 0.
\end{align*}$$
This makes it clear that if $a$ satisfies $f_n(a) = 0$, then $g_n(a) = 0$. I suppose then that the solution to the question is dependent on if the sequence $g_n$ defined by $g_0 = x$ and 
$$g_n = g_{n-1}^2 - x - n$$
has an explicit form.
