TITLE: This infinitely nested root gives me two answers $ \sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{...}}}}}} $
QUESTION [30 upvotes]: I am trying to evaluate
$$ \sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{...}}}}}} $$
where those numbers inside roots are
$$ a_{n+1}=\frac{a_n^2}{2}$$
And I found two ways to solve it that give different answers. I believe one of those is not right, but I don't know which and why. Please help.
Method-1.
$$x+1=\sqrt{x^2+2x+1}=\sqrt{x^2+x+\sqrt{x^2+2x+1}}\\
=\sqrt{x^2+x+\sqrt{x^2+x+\sqrt{x^2+2x+1}}}=...$$
$x=1\rightarrow$
$$2=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}}$$
Therefore
$$\begin{align}
&4=2\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}}\\
&=\sqrt{2^2\cdot2+2^2\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}}\\
&=\sqrt{8+\sqrt{2^4\cdot2+2^4\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{...}}}}}}\\
&=\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{...}}}}}\end{align}$$
Finally,
$$\sqrt{4+4}=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{...}}}}}}=2\sqrt2$$
Method-2.
$$\begin{align}
&x+2=\sqrt{x^2+4x+4}=\sqrt{x^2+3x+\sqrt{x^2+8x+16}}\\
&=\sqrt{x^2+3x+\sqrt{x^2+7x+\sqrt{x^2+32x+256}}}\\
&=\sqrt{x^2+3x+\sqrt{x^2+7x+\sqrt{x^2+31x+\sqrt{x^2+512x+256^2}}}}...
\end{align}$$
$x=1\rightarrow$
$$3=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{...}}}}}} $$
Alternative Method-2.
$$\begin{align}
&3=\sqrt9=\sqrt{4+5}=\sqrt{4+\sqrt{25}}=\sqrt{4+\sqrt{8+17}}\\
&=\sqrt{4+\sqrt{8+\sqrt{2\cdot16+16^2+1}}}\\
&=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{2\cdot16^2+16^4+1}}}}\\
&=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{2\cdot16^4+16^8+1}}}}}\\
&=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{2\cdot16^8+16^{16}+1}}}}}}=...
\end{align}$$
So, I have two answers $2\sqrt2$ and $3$. Which one is correct and what's the problem in the other solution?? Thanks.

Now I think I understand. Thanks for all the answers.
Let me post this method-3 just to show that it could be any number $\geq2\sqrt2$ and conclude this topic.
Method-3.
$$\begin{align}
&\sqrt{10}=\sqrt{4+6}=\sqrt{4+\sqrt{36}}=\sqrt{4+\sqrt{8+28}}\\
&=\sqrt{4+\sqrt{8+\sqrt{32+752}}}\\
&=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+564992}}}}\\
&=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{...}}}}}}=...
\end{align}$$

REPLY [6 votes]: As Jason said in the comments, infinite radicals are not well defined. Let me show that you can in fact get any $x\geq 2\sqrt{2}$ using an infinite radicals looking like yours.
In other words, for any $x$, there exist a sequence $(c_n)$ such that
$$x=\sqrt{4+\sqrt{8+\sqrt{32+...+\sqrt{2^{\alpha_n}+c_n}}}}$$
where $\alpha_n=2^n+1$ (your sequence if I'm not wrong).
Well, construct the sequence $(c_n)$ inductively. Put $c_0=x^2-4$, so that
$x=\sqrt{4+c_0}$.
Then $c_1=(x^2-4)^2-8$ and so on.
The particular thing with $2\sqrt{2}$ is that this is the limit of the sequence $(u_n)$ where
$$u_n=\sqrt{4+\sqrt{8+...+\sqrt{2^{\alpha_n}}}}$$
so that if $x\geq 2\sqrt{2}$, the $c_i$'s will always be positive, so that the sequence $(c_i)$ will be well defined.