<s>[INST] 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
$$\begin{align}
f_n(x) &= \sqrt{n^x + \sqrt{n^{x+1} + \sqrt{n^{x+2} + \sqrt{n^{x+3} + \ldots}}}} \\
f_n(x)^2 &= n^x + f_n(x + 1)~\text ?
\end{align}$$
 [/INST] 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.
</s>