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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.

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

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  • $\begingroup$ Note that it should be $\sqrt{25}$ not $\sqrt25$ on the first line of alternative method 2. I tried to edit this, but it said I had to change at least six characters, but I only needed to add 2! $\endgroup$ – Sam T Nov 27 '15 at 18:55
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    $\begingroup$ @Smiley Sam Thanks, I edited it. $\endgroup$ – Kay K. Nov 27 '15 at 18:57
  • $\begingroup$ I can't find the mistake though! Been looking for a while! $\endgroup$ – Sam T Nov 27 '15 at 18:57
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    $\begingroup$ The infinite square root isn't really well-defined. The most natural definition is the limit of a certain sequence (in this case, $(\sqrt{4},\sqrt{4+\sqrt{8}},\sqrt{4+\sqrt{8+\sqrt{32}}},\ldots)$) but one must be careful about convergence when discussing the limit. It's possible that, by dealing with this problem in an intuitive rather than rigorous way, you are implicitly using different definitions. $\endgroup$ – Jason Nov 27 '15 at 19:03
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    $\begingroup$ Are you sure the sequence converges? My intuition says that it diverges. $\endgroup$ – Brady Gilg Nov 27 '15 at 19:35
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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.

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  • $\begingroup$ Thanks. One question still bothers me is that what is still with $3$ then? I tried to make another sequence that also looks like one of these and converges to something else other than $2\sqrt2$ ot $3$, but I was not successful. $2\sqrt2$ may be the special one, but what's specian with $3$? Shouldn't I be able to make another sequence that converges to for example 4? But I can do that only for $2\sqrt2$ and $3$... $\endgroup$ – Kay K. Dec 2 '15 at 3:47
  • $\begingroup$ Never mind. I think it was just a misconception that $3$ looked special. $\endgroup$ – Kay K. Dec 2 '15 at 4:18
  • $\begingroup$ @KayK. in fact all the positive sequences can be explicitly given as $(2^{\alpha 2^n} + 2^{-\alpha 2^n})2^{2^{n-1}}$. $\alpha = 0$ corresponds to the sequence starting with $2\sqrt 2$, $\alpha = 0.5$ is the one starting with $3$, and applying $\phi$ corresponds to doubling $\alpha$, So yeah $3$ doesn't look that special among those. $\endgroup$ – mercio Dec 2 '15 at 14:07
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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.

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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.

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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.

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    $\begingroup$ Thanks but, I can say the same thing for the first sequence. $\sqrt8=\sqrt{4+4}=\sqrt{4+\sqrt{16}}=\sqrt{4+\sqrt{8+\sqrt{64}}}=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{1024}}}}=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{512^2}}}}}=\sqrt{4+\sqrt{8+\sqrt{32+\sqrt{512+\sqrt{512^2/2+\sqrt{(512^2/2)^2}}}}}}=...$ which is just $\sqrt8$,$\sqrt8$,$\sqrt8$,... $\endgroup$ – Kay K. Nov 28 '15 at 0:56
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    $\begingroup$ Both methods are not rigorous. It is just a coincidence that you got the right answer the first way. You should first show the infinite radical converges and then use an argument like bburGsamohT's to conclude what the limit is. $\endgroup$ – ET93 Nov 30 '15 at 0:31
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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.

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