This infinitely nested root gives me two answers $ \sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{...}}}}}} $

Question

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.

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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.

$$\begin{align} &\sqrt{10}=\sqrt{4+6}=\sqrt{4+\sqrt{36}}=\sqrt{4+\sqrt{8+28}}\\ &=\sqrt{4+\sqrt{8+\sqrt{32+752}}}\\ &=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+564992}}}}\\ &=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{\frac{512^2}{2}+\sqrt{...}}}}}}=... \end{align}$$

Answer 1

Your method $2$ is merely an observation that the sequence $a_n = 2^{2^n}+1$ is positive and satisfies the recurrence relation $a_{n+1} = a_n^2 - 2^{2^n+1}$.

The sequence $a_n = 2^{2^{n-1}+1}$ is another such sequence, and in fact it is the smallest sequence that stays positive forall $n$ among the sequences satisfying the relation.

Let $S_n$ be the set of values of $a_0$ such that $a_0,a_1,\ldots,a_n \ge 0$, and $S$ be the intersection of all those sets. It is pretty clear that $S_n$ is of the form $[x_n ; \infty)$ for some increasing sequence $(x_n)$ and that $x_n \le 2 \sqrt 2$ (because $2\sqrt 2$ engenders a positive sequence).

If you look carefully, $\sqrt 4$ is the value of $a_0$ making $a_1=0$, so it is $x_1$, the minimum of $S_1$. Likewise, $\sqrt {4+\sqrt 8}$ is $x_2$ the minimum of $S_2$, etc. So the sequence of finite nests is the sequence $\min(S_n)$. So the infinite nested expression is the limit of those values, which means the lower bound of $S$. (and in fact it's a minimum again)

Now your first method is close to saying that if $(a_n)$ is a positive sequence satisfying the recurrence relation, then the sequence $(b_n)$ defined by $b_n = a_{n+1} 2^{-2^{n-1}}$ also satisfies the recurrence relation and is also positive forall $n$.
This shows that if $\phi(a) = (a^2-4)/\sqrt 2$, then $S$ is stable by $\phi$.

Now, $2\sqrt 2 \in S$ and is a repulsive fixpoint of $\phi$. It turns out that if you iterate $\phi$ on any $a < 2\sqrt 2$ you eventually land on a negative value, and since $S$ doesn't have negative values, they can't belong to $S$.

So $2\sqrt 2$ is the minimum of $S$, and thus it is the value of the limit of the nested square roots.

Answer 2

Let $$x=\sqrt{4+\sqrt{8+\dots}}$$ Then $$x^2=4+\sqrt{8+\dots}=4+\sqrt{2}\cdot{x}.$$ So $x$ satisfies $x^2-\sqrt{2}\cdot x-4=0$, i.e. the first solution is correct. I am not sure off the top of my head what is wrong with the second one, but I will keep looking.

Answer 3

I think the problem is that the sequences you are taking limits of are different in the two methods. In the first method, you take the sequence $\sqrt{4},\sqrt{4+\sqrt{8}}$,... This is the sequence implicitly behind the expression $\sqrt{4+\sqrt{8+\sqrt{32+...}}}$. On the other hand, in your other method, you look at the sequence $\sqrt{4+5}, \sqrt{4+\sqrt{8+17}}, ...$ which is just $3,3,3, ...$ There is nothing wrong with different sequences having different limits.

Answer 4

Suppose you have two integer sequences, $\{a_n\}$ and $\{b_n\}$. I believe the crux of the apparent paradox lies in the observation that if one defines

$$ x = \sqrt{a_1+\sqrt{a_2+\sqrt{a_3+\sqrt{a_4+\cdots}}}} $$

as the limit of the sequence

$$ l_1 = \sqrt{a_1} $$ $$ l_2 = \sqrt{a_1+\sqrt{a_2}} $$ $$ l_3 = \sqrt{a_1+\sqrt{a_2+\sqrt{a_3}}} $$

et cetera, one may end up with a different result than if one defines it as the limit of the sequence

$$ m_1 = \sqrt{a_1+b_1} $$ $$ m_2 = \sqrt{a_1+\sqrt{a_2+b_2}} $$ $$ m_3 = \sqrt{a_1+\sqrt{a_2+\sqrt{a_3+b_3}}} $$

et cetera. In particular, if the $\{b_n\}$ converge, or diverge gently, then the two sequences should have the same limit (if indeed it exists). My intuition is that with the nested square roots as they are, if the $\{b_n\}$ diverges in the limit as $2^{2^n}$, or faster, then the two limits (if they both exist) will be different, but I'm not confident of that.

Answer 5

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.