Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$ Inspired by Ramanujan's problem and solution of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$, I decided to attempt evaluating the infinite radical
$$
\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
$$
Taking a cue from Ramanujan's solution method, I defined a function $f(x)$ such that
$$
f(x) = \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}}
$$
We can see that
$$\begin{align}
f(0) &= \sqrt{2^0 + \sqrt{2^1 + \sqrt{2^2 + \sqrt{2^3 + \ldots}}}} \\
&= \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
\end{align}$$
And we begin solving by
$$\begin{align}
f(x) &= \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}} \\
f(x)^2 &= 2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}} \\
&= 2^x + f(x + 1) \\
f(x + 1) &= f(x)^2 - 2^x
\end{align}$$
At this point I find myself stuck, as I have little experience with recurrence relations.
How would this recurrence relation be solved? Would the method extend easily to
$$\begin{align}
f_n(x) &= \sqrt{n^x + \sqrt{n^{x+1} + \sqrt{n^{x+2} + \sqrt{n^{x+3} + \ldots}}}} \\
f_n(x)^2 &= n^x + f_n(x + 1)~\text ?
\end{align}$$
 A: Let $x_0 = \sqrt{1}$, $x_1 = \sqrt{1 + \sqrt{2}}$, $x_3 = \sqrt{1 + \sqrt{2 + \sqrt{4}}}$, and so on. Then we have:
$$\sqrt{1 + \sqrt{2 + \sqrt{4 + \cdots}}} = \lim_{n \to \infty} x_n$$
Clearly, this sequence is monotonically increasing. It also converges, since we can see each new term appears under $n$ square roots, and hence $| x_n - x_{n - 1} | \propto 2^{-n}$ which should be enough.

From observation, the value of $x_n - x_{n - 1}$ is a root of a polynomial of order $2^{2n - 3}$. In this sense, a closed-form solution is very unlikely to exist. Not a full answer, but it looks complicated.
A: Introduce the notation $[a_0]=\sqrt{a_0}$, and $[a_0,a_1]=\sqrt{a_0+\sqrt{a_1}}$, and so on, including infinite lists:
$$[a_0,a_1,a_2,...]=\sqrt{a_0 + \sqrt{a_1 + \sqrt{a_2 + \cdots}}}=\sqrt{a_0+[a_1,a_2,\ldots]}.$$
Generally $[a_0,a_1,\ldots]^2 = a_0 + [a_1,a_2,\ldots]$, so for constant-term lists we have a closed-form solution:
$$
[x,x,\ldots]^2=x+[x,x,\ldots] \implies [x,x,\ldots]=\frac{1}{2}+\frac{1}{2}\sqrt{1+4x}.
$$
If $b_i \le a_i$ for each $i$, then clearly $[b_0,b_1,\ldots]\le[a_0,a_1,\ldots]$.  What happens when a multiplicative factor is introduced?  You have
$$k[a_0,a_1,a_2,...]=\sqrt{k^2 a_0 + k^2[a_1,a_2,...]}=[k^2a_0,k^4a_1,k^8a_2,\ldots]$$
In your case,
$[1,2,4,\ldots]=\sqrt{1+[2,4,8,\ldots]}=\sqrt{1+\sqrt{2}[1,1,1/2,1/16,\ldots]}$.  Using the bounds
$$
\sqrt{2}=[1,1]\le[1,1,1/2,1/16,\ldots]\le[1,1,1,\ldots]=\frac{1}{2}+\frac{1}{2}\sqrt{5},
$$
you have
$$
1.732 \approx \sqrt{3} \le [1,2,4,\ldots] \le \sqrt{1+\frac{1+\sqrt{5}}{\sqrt{2}}}\approx 1.813.
$$
Tighter bounds can be provided, of course, but this suffices to show that the limit exists.
A: Just a comment on such similar cases. From this short answer,
$$x+2^n=\sqrt{4^{n}+x\sqrt{4^{\left(n+1\right)}+x\sqrt{4^{\left(n+2\right)}+x\sqrt{...}}}}\tag{1}$$
Let $N=\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$. Then
$$\begin{aligned}
 N &< \sqrt{1+\sqrt{2+\sqrt{4^{1}+\sqrt{4^{2}+\sqrt{4^{3}+...}}}}} \\
 N &< \sqrt{1+\sqrt{5}} \\
  &< 1.79890743995\\
\end{aligned}$$
For comparison, $N ≃ 1.78316580926410..$. Also from Eq. $1$, we can get;
$$\boxed{1+\sqrt{2}=\sqrt{2^{1}+\sqrt{2^{3}+\sqrt{2^{5}+\sqrt{2^{7}...}}}}}$$
$$\begin{aligned}
N &< \sqrt{1+\sqrt{2^{1}+\sqrt{2^{3}+\sqrt{2^{5}+\sqrt{2^{7}...}}}}} \\
 &< \sqrt{2+\sqrt{2}} \\
\end{aligned}$$
A: Since you have recurrence relation $(f(x))^2=2^x+f(x+1)$ you could find an approximate solution by approximating $f(x+1)\approx f(x)$ and then you get quadratic equation for the function $f$ and in doing so you can find an approximate value $f(x)$ for every value of $x$. It is clear from the recurrence relation that the exact expression for $f(x)$ may not be expressible as some combination of the functions that have been studied to this day, but maybe I am wrong.
If you still want to search for closed forms for certain values of $f$ maybe it is better to study expressions of this type that are finite, such as $f(2,x,n) =\sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} +\sqrt {\ldots+\sqrt{2^{x+n}}}}}}}$ and then let $n\to\infty$ to get your $f$.
A: This is not an answer to your question, but you may be interested in the following two similar evaluations:
$$\sqrt{1+\sqrt{4+\sqrt{16+\sqrt{64+...}}}}=2\tag{1}$$
$$\sqrt{1+2^{-1}\sqrt{1+2^{-2}\sqrt{1+2^{-3}\sqrt{1+...}}}}=\frac{5}{4}\tag{2}$$
To prove $(1)$, one may use the fact that
$$2^n+1=\sqrt{4^n+2^{n+1}+1}$$
and $(2)$ can be obtained from $(1)$ by multiplying it by $1/2$ and inverting the first radical.
A: I'd rather try this:
If
$$x=\sqrt{ 1+\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}},$$
then 
$$\begin{align}ux&=u\sqrt{ 1+\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+u^2\sqrt{2+\sqrt{4+\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+\sqrt{2u^4+u^4\sqrt{4+\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+\sqrt{2u^4+\sqrt{4u^8+u^8\sqrt{8+\ldots}}}}\\
&=\sqrt{ u^2+\sqrt{2u^4+\sqrt{4u^8+\sqrt{8u^{16}+\ldots}}}}\\
\end{align}$$
Thus if $u=\frac1{\sqrt 2}$ then 
$$ \frac x{\sqrt2}=\sqrt{\frac12+\sqrt{\frac12+\sqrt{\frac12+\sqrt{\frac12+\ldots}}}}$$
The right hand side $y$ has the property that $y^2-\frac12=y$, i.e. can be found by solving a quadratic.
