What is the maximum value of this nested radical?
Clash Royale CLAN TAG#URR8PPP
I was experimenting on Desmos (as usual), in particular infinite recursions and series. Here is one that was of interest:
What is the maximum value of $$F_infty=sqrt{frac{x}{x+sqrt{frac{x^2}{xsqrt{frac{x^3}{x+ sqrt{ frac{x^4}{xcdots}}}}}}}}$$ where the sign alternates and the power in each numerator increases by one?
Some observations follow.

Let $$F_k=underbrace{sqrt{frac{x}{x+sqrt{frac{x^2}{xsqrt{frac{x^3}{x+ sqrt{ frac{x^4}{xsqrt{fraccdots{xpmsqrt{x^k}}}}}}}}}}}}_{k,text{times}}.$$ For large nests, say after $k=10$, the function monotonically increases from zero onwards. It is hopeless to simply rearrange $F_infty$ since the powers increase each time – we can no longer write $F_infty$ as a function of itself to be solved.

Here is a plot of $F_{15}$:
 What is striking is that the largest value of $x$ in the domain of $F_k$ decreases as $k$ increases. Based on the plot, I think that the domain of $F_infty$ is $[0,1]$. This is because for large $x$, the denominator of the square roots will be larger than its successor, which is absurd as we are working only in $Bbb R$.
 Furthermore, I also conjecture that $$max F_infty=phi1=frac{sqrt51}2,$$ the positive solution of the equation $x^2+x1=0$. This seems right as $max F_{15}=0.6179$ from the plot.
I’d appreciate proofs to my thoughts, especially the last one!
$endgroup$
This question has an open bounty worth +100
reputation from TheSimpliFire ending in 5 days.
The current answers do not contain enough detail.
This bounty will be awarded to whoever manages to prove that $F_infty$ is monotonically increasing in the domain $[0,1]$. This is the last part to showing that $max F_infty = phi – 1$.
add a comment 
I was experimenting on Desmos (as usual), in particular infinite recursions and series. Here is one that was of interest:
What is the maximum value of $$F_infty=sqrt{frac{x}{x+sqrt{frac{x^2}{xsqrt{frac{x^3}{x+ sqrt{ frac{x^4}{xcdots}}}}}}}}$$ where the sign alternates and the power in each numerator increases by one?
Some observations follow.

Let $$F_k=underbrace{sqrt{frac{x}{x+sqrt{frac{x^2}{xsqrt{frac{x^3}{x+ sqrt{ frac{x^4}{xsqrt{fraccdots{xpmsqrt{x^k}}}}}}}}}}}}_{k,text{times}}.$$ For large nests, say after $k=10$, the function monotonically increases from zero onwards. It is hopeless to simply rearrange $F_infty$ since the powers increase each time – we can no longer write $F_infty$ as a function of itself to be solved.

Here is a plot of $F_{15}$:
 What is striking is that the largest value of $x$ in the domain of $F_k$ decreases as $k$ increases. Based on the plot, I think that the domain of $F_infty$ is $[0,1]$. This is because for large $x$, the denominator of the square roots will be larger than its successor, which is absurd as we are working only in $Bbb R$.
 Furthermore, I also conjecture that $$max F_infty=phi1=frac{sqrt51}2,$$ the positive solution of the equation $x^2+x1=0$. This seems right as $max F_{15}=0.6179$ from the plot.
I’d appreciate proofs to my thoughts, especially the last one!
$endgroup$
This question has an open bounty worth +100
reputation from TheSimpliFire ending in 5 days.
The current answers do not contain enough detail.
This bounty will be awarded to whoever manages to prove that $F_infty$ is monotonically increasing in the domain $[0,1]$. This is the last part to showing that $max F_infty = phi – 1$.

1$begingroup$
How exactly does the expression for $F_k$ look? I can’t actually work out easily how it terminates at the k^th step.
$endgroup$
– T_M
Jan 5 at 17:20
add a comment 
I was experimenting on Desmos (as usual), in particular infinite recursions and series. Here is one that was of interest:
What is the maximum value of $$F_infty=sqrt{frac{x}{x+sqrt{frac{x^2}{xsqrt{frac{x^3}{x+ sqrt{ frac{x^4}{xcdots}}}}}}}}$$ where the sign alternates and the power in each numerator increases by one?
Some observations follow.

Let $$F_k=underbrace{sqrt{frac{x}{x+sqrt{frac{x^2}{xsqrt{frac{x^3}{x+ sqrt{ frac{x^4}{xsqrt{fraccdots{xpmsqrt{x^k}}}}}}}}}}}}_{k,text{times}}.$$ For large nests, say after $k=10$, the function monotonically increases from zero onwards. It is hopeless to simply rearrange $F_infty$ since the powers increase each time – we can no longer write $F_infty$ as a function of itself to be solved.

Here is a plot of $F_{15}$:
 What is striking is that the largest value of $x$ in the domain of $F_k$ decreases as $k$ increases. Based on the plot, I think that the domain of $F_infty$ is $[0,1]$. This is because for large $x$, the denominator of the square roots will be larger than its successor, which is absurd as we are working only in $Bbb R$.
 Furthermore, I also conjecture that $$max F_infty=phi1=frac{sqrt51}2,$$ the positive solution of the equation $x^2+x1=0$. This seems right as $max F_{15}=0.6179$ from the plot.
I’d appreciate proofs to my thoughts, especially the last one!
$endgroup$
I was experimenting on Desmos (as usual), in particular infinite recursions and series. Here is one that was of interest:
What is the maximum value of $$F_infty=sqrt{frac{x}{x+sqrt{frac{x^2}{xsqrt{frac{x^3}{x+ sqrt{ frac{x^4}{xcdots}}}}}}}}$$ where the sign alternates and the power in each numerator increases by one?
Some observations follow.

Let $$F_k=underbrace{sqrt{frac{x}{x+sqrt{frac{x^2}{xsqrt{frac{x^3}{x+ sqrt{ frac{x^4}{xsqrt{fraccdots{xpmsqrt{x^k}}}}}}}}}}}}_{k,text{times}}.$$ For large nests, say after $k=10$, the function monotonically increases from zero onwards. It is hopeless to simply rearrange $F_infty$ since the powers increase each time – we can no longer write $F_infty$ as a function of itself to be solved.

Here is a plot of $F_{15}$:
 What is striking is that the largest value of $x$ in the domain of $F_k$ decreases as $k$ increases. Based on the plot, I think that the domain of $F_infty$ is $[0,1]$. This is because for large $x$, the denominator of the square roots will be larger than its successor, which is absurd as we are working only in $Bbb R$.
 Furthermore, I also conjecture that $$max F_infty=phi1=frac{sqrt51}2,$$ the positive solution of the equation $x^2+x1=0$. This seems right as $max F_{15}=0.6179$ from the plot.
I’d appreciate proofs to my thoughts, especially the last one!
This question has an open bounty worth +100
reputation from TheSimpliFire ending in 5 days.
The current answers do not contain enough detail.
This bounty will be awarded to whoever manages to prove that $F_infty$ is monotonically increasing in the domain $[0,1]$. This is the last part to showing that $max F_infty = phi – 1$.
This question has an open bounty worth +100
reputation from TheSimpliFire ending in 5 days.
The current answers do not contain enough detail.
This bounty will be awarded to whoever manages to prove that $F_infty$ is monotonically increasing in the domain $[0,1]$. This is the last part to showing that $max F_infty = phi – 1$.

1$begingroup$
How exactly does the expression for $F_k$ look? I can’t actually work out easily how it terminates at the k^th step.
$endgroup$
– T_M
Jan 5 at 17:20
add a comment 

1$begingroup$
How exactly does the expression for $F_k$ look? I can’t actually work out easily how it terminates at the k^th step.
$endgroup$
– T_M
Jan 5 at 17:20
How exactly does the expression for $F_k$ look? I can’t actually work out easily how it terminates at the k^th step.
$endgroup$
– T_M
Jan 5 at 17:20
How exactly does the expression for $F_k$ look? I can’t actually work out easily how it terminates at the k^th step.
$endgroup$
– T_M
Jan 5 at 17:20
add a comment 
1 Answer
1
active
oldest
votes
Zachary’s and OP’s combined solutions
If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_infty$ at $x=1$ will give the maximum, which will be an infinite fraction:
$$F_infty (1) = sqrt{ frac{1}{1+sqrt{ frac{1}{1 sqrt{cdots} }} }} = sqrt{ frac{1}{1+sqrt{ frac{1}{1 F_infty (1) }} }}implies F_infty (1)^2 left(1+sqrt{ frac{1}{1 F_infty (1) }} right) = 1$$ so $$(F_infty(1)^21)^2=frac{F_infty(1)^4}{1F_infty(1)}implies F_infty(1)^52F_infty(1)^3+2F_infty(1)^2+F_infty(1)1=0.$$ Factoring out some roots, we get $$(F_infty(1)^2+F_infty(1)1)(F_infty(1)^3F_infty(1)^2+1)=0$$ It can be verified from WA, for example, that the only positive real solution is at $F_infty(1)=phi1$ derived from the first quadratic factor.
Proof attempt for the domain:
While I’m not sure how this holds up for $F_{infty}(x)$, we can show that $forall x>1$ $exists y mid forall n geq y, F_n(x) notin mathbb{R}$.
Now, if we ever get a negative denominator, the end result will be nonreal. This is because addition, subtraction, and division between nonreals and nonzero reals will stay nonreal, and the square root of a nonreal will also be nonreal.
Now, when $n>2$ is odd, $sqrt{x^n}<x$, and thus $F_n(x) notin mathbb{R}$. Thus, we must concern ourselves with even $n$.
So, in $F_{2k}(x)$, consider:
$$ frac{cdots}{x – sqrt{ frac{x^{2k1}}{x+sqrt{x^{2k}}} }}$$
$$ x < sqrt{ frac{x^{2k1}}{x+x^k} } implies F_{2k}(x) notin mathbb{R} $$
Now, we will show that the “bottom” of $F_{2(k+1)}(x)$ is greater than the bottom of $F_{2k}(x)$, for $x>1$:
$$ sqrt{ frac{x^{2k1}}{x+sqrt{x^{2k}}} } < sqrt{ frac{x^{2k+1}}{x+sqrt{x^{2(k+1)}}} } $$
$$ frac{x^{2k1}}{x+x^k} < frac{x^{2k+1}}{x+x^{k+1}} $$
$$ x^{2k}+x^{3k} < x^{2k+2} + x^{3k+2} $$
The bottom inequality is true for $x>1$, thus by reversing the work, we prove the first line to be true. (we could also do this with partial derivatives, but that’s messier in my opinion.
Using this result, it follows that:
$$ x < sqrt{ frac{x^{2k1}}{x+x^k} } implies x < sqrt{ frac{x^{2k+1}}{x+x^{k+1}} } implies F_{2(k+1)}(x) notin mathbb{R} $$
Now, to prove the “bottom” will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $x > 1$. Since $x$ is nonnegative, the domain is $[0,1]$.
So, for any $x$ there’s an $y$ for which $F_n(x)$ is nonreal for finite $n$ greater than $y$. However, I’m not certain this is rigorously extends to the infinite case.
Extra notes: I’ve rigged up a computer program to calculate when $F_n(x)$ becomes nonreal. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+frac{1}{2^k}$.
$$smallbegin{array}{cc} k&0&1&2&3&4&5&6&7&8&9&10&11&12&13&14\hline n_{text{diverge}}&8&8&10&14&22&38&68&130&252&500&992&1978&3948&7890&15774end{array}$$
As you can see, as $x$ gets twice as close to $1$, it takes almost twice as many terms to go nonreal. I’ve tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $frac{1}{x1}$.
Proof attempt for monotonicity: (next improvement is to prove that $H'<1$)
Here I will attempt to prove the monotonicity of $F_infty$. First, let us introduce some definitions. $$F:=F_{infty},quad G:=sqrt{frac{x^2}{xsqrt{frac{x^3}{x+sqrt{frac{x^4}{x sqrt{ frac{x^5}{x+cdots}}}}}}}},quad H:=sqrt{frac{x^3}{x+sqrt{frac{x^4}{xsqrt{frac{x^5}{x+ sqrt{ frac{x^6}{x+cdots}}}}}}}}$$ Since $F=sqrt{dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F’=frac1{2F}cdotfrac{1(x+G)x(1+G’)}{(x+G)^2}>0impliedby GxG’>0$$ as $(x+G)^2$ and $F$ are clearly nonnegative.
Now this is implied by $$G’=frac1{2G}cdotleft(1+frac{H’x^2H}{(xH)^2}right)<frac Gx$$ and since $G=sqrt{dfrac{x^2}{xH}}$ (note that $H<x$), we get begin{align}2G^2>x+xfrac{H’x^2H}{(xH)^2}&impliedbyfrac{2x}{xH}>1+frac{H’x^2H}{(xH)^2}\&impliedby 2x^22Hx>x^22Hx+H^2+H’x^2H^2\&impliedby x^2>H’x^2impliedby H'<1end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H’$. Of course, the latter two are only close approximations to the real distribution of $H$.
$endgroup$

1$begingroup$
The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.
$endgroup$
– Zachary Hunter
Jan 5 at 18:18 
$begingroup$
Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.
$endgroup$
– Zachary Hunter
Jan 7 at 0:43 
1$begingroup$
Well done! It’s not too rigorous, but I can understand your method. I’ll make some edits later. I have made an attempt into showing its monotonicity, but I can’t get any farther than $H'<1$. Any help appreciated!
$endgroup$
– TheSimpliFire
2 days ago
add a comment 
Your Answer
StackExchange.ifUsing(“editor”, function () {
return StackExchange.using(“mathjaxEditing”, function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [[“$”, “$”], [“\\(“,”\\)”]]);
});
});
}, “mathjaxediting”);
StackExchange.ready(function() {
var channelOptions = {
tags: “”.split(” “),
id: “69”
};
initTagRenderer(“”.split(” “), “”.split(” “), channelOptions);
StackExchange.using(“externalEditor”, function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using(“snippets”, function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: ‘answer’,
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: “”,
imageUploader: {
brandingHtml: “Powered by u003ca class=”iconimgurwhite” href=”https://imgur.com/”u003eu003c/au003e”,
contentPolicyHtml: “User contributions licensed under u003ca href=”https://creativecommons.org/licenses/bysa/3.0/”u003ecc bysa 3.0 with attribution requiredu003c/au003e u003ca href=”https://stackoverflow.com/legal/contentpolicy”u003e(content policy)u003c/au003e”,
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: “.discardanswer”
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave(‘#loginlink’);
});
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin(‘.newpostlogin’, ‘https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062908%2fwhatisthemaximumvalueofthisnestedradical%23newanswer’, ‘question_page’);
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Zachary’s and OP’s combined solutions
If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_infty$ at $x=1$ will give the maximum, which will be an infinite fraction:
$$F_infty (1) = sqrt{ frac{1}{1+sqrt{ frac{1}{1 sqrt{cdots} }} }} = sqrt{ frac{1}{1+sqrt{ frac{1}{1 F_infty (1) }} }}implies F_infty (1)^2 left(1+sqrt{ frac{1}{1 F_infty (1) }} right) = 1$$ so $$(F_infty(1)^21)^2=frac{F_infty(1)^4}{1F_infty(1)}implies F_infty(1)^52F_infty(1)^3+2F_infty(1)^2+F_infty(1)1=0.$$ Factoring out some roots, we get $$(F_infty(1)^2+F_infty(1)1)(F_infty(1)^3F_infty(1)^2+1)=0$$ It can be verified from WA, for example, that the only positive real solution is at $F_infty(1)=phi1$ derived from the first quadratic factor.
Proof attempt for the domain:
While I’m not sure how this holds up for $F_{infty}(x)$, we can show that $forall x>1$ $exists y mid forall n geq y, F_n(x) notin mathbb{R}$.
Now, if we ever get a negative denominator, the end result will be nonreal. This is because addition, subtraction, and division between nonreals and nonzero reals will stay nonreal, and the square root of a nonreal will also be nonreal.
Now, when $n>2$ is odd, $sqrt{x^n}<x$, and thus $F_n(x) notin mathbb{R}$. Thus, we must concern ourselves with even $n$.
So, in $F_{2k}(x)$, consider:
$$ frac{cdots}{x – sqrt{ frac{x^{2k1}}{x+sqrt{x^{2k}}} }}$$
$$ x < sqrt{ frac{x^{2k1}}{x+x^k} } implies F_{2k}(x) notin mathbb{R} $$
Now, we will show that the “bottom” of $F_{2(k+1)}(x)$ is greater than the bottom of $F_{2k}(x)$, for $x>1$:
$$ sqrt{ frac{x^{2k1}}{x+sqrt{x^{2k}}} } < sqrt{ frac{x^{2k+1}}{x+sqrt{x^{2(k+1)}}} } $$
$$ frac{x^{2k1}}{x+x^k} < frac{x^{2k+1}}{x+x^{k+1}} $$
$$ x^{2k}+x^{3k} < x^{2k+2} + x^{3k+2} $$
The bottom inequality is true for $x>1$, thus by reversing the work, we prove the first line to be true. (we could also do this with partial derivatives, but that’s messier in my opinion.
Using this result, it follows that:
$$ x < sqrt{ frac{x^{2k1}}{x+x^k} } implies x < sqrt{ frac{x^{2k+1}}{x+x^{k+1}} } implies F_{2(k+1)}(x) notin mathbb{R} $$
Now, to prove the “bottom” will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $x > 1$. Since $x$ is nonnegative, the domain is $[0,1]$.
So, for any $x$ there’s an $y$ for which $F_n(x)$ is nonreal for finite $n$ greater than $y$. However, I’m not certain this is rigorously extends to the infinite case.
Extra notes: I’ve rigged up a computer program to calculate when $F_n(x)$ becomes nonreal. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+frac{1}{2^k}$.
$$smallbegin{array}{cc} k&0&1&2&3&4&5&6&7&8&9&10&11&12&13&14\hline n_{text{diverge}}&8&8&10&14&22&38&68&130&252&500&992&1978&3948&7890&15774end{array}$$
As you can see, as $x$ gets twice as close to $1$, it takes almost twice as many terms to go nonreal. I’ve tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $frac{1}{x1}$.
Proof attempt for monotonicity: (next improvement is to prove that $H'<1$)
Here I will attempt to prove the monotonicity of $F_infty$. First, let us introduce some definitions. $$F:=F_{infty},quad G:=sqrt{frac{x^2}{xsqrt{frac{x^3}{x+sqrt{frac{x^4}{x sqrt{ frac{x^5}{x+cdots}}}}}}}},quad H:=sqrt{frac{x^3}{x+sqrt{frac{x^4}{xsqrt{frac{x^5}{x+ sqrt{ frac{x^6}{x+cdots}}}}}}}}$$ Since $F=sqrt{dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F’=frac1{2F}cdotfrac{1(x+G)x(1+G’)}{(x+G)^2}>0impliedby GxG’>0$$ as $(x+G)^2$ and $F$ are clearly nonnegative.
Now this is implied by $$G’=frac1{2G}cdotleft(1+frac{H’x^2H}{(xH)^2}right)<frac Gx$$ and since $G=sqrt{dfrac{x^2}{xH}}$ (note that $H<x$), we get begin{align}2G^2>x+xfrac{H’x^2H}{(xH)^2}&impliedbyfrac{2x}{xH}>1+frac{H’x^2H}{(xH)^2}\&impliedby 2x^22Hx>x^22Hx+H^2+H’x^2H^2\&impliedby x^2>H’x^2impliedby H'<1end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H’$. Of course, the latter two are only close approximations to the real distribution of $H$.
$endgroup$

1$begingroup$
The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.
$endgroup$
– Zachary Hunter
Jan 5 at 18:18 
$begingroup$
Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.
$endgroup$
– Zachary Hunter
Jan 7 at 0:43 
1$begingroup$
Well done! It’s not too rigorous, but I can understand your method. I’ll make some edits later. I have made an attempt into showing its monotonicity, but I can’t get any farther than $H'<1$. Any help appreciated!
$endgroup$
– TheSimpliFire
2 days ago
add a comment 
Zachary’s and OP’s combined solutions
If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_infty$ at $x=1$ will give the maximum, which will be an infinite fraction:
$$F_infty (1) = sqrt{ frac{1}{1+sqrt{ frac{1}{1 sqrt{cdots} }} }} = sqrt{ frac{1}{1+sqrt{ frac{1}{1 F_infty (1) }} }}implies F_infty (1)^2 left(1+sqrt{ frac{1}{1 F_infty (1) }} right) = 1$$ so $$(F_infty(1)^21)^2=frac{F_infty(1)^4}{1F_infty(1)}implies F_infty(1)^52F_infty(1)^3+2F_infty(1)^2+F_infty(1)1=0.$$ Factoring out some roots, we get $$(F_infty(1)^2+F_infty(1)1)(F_infty(1)^3F_infty(1)^2+1)=0$$ It can be verified from WA, for example, that the only positive real solution is at $F_infty(1)=phi1$ derived from the first quadratic factor.
Proof attempt for the domain:
While I’m not sure how this holds up for $F_{infty}(x)$, we can show that $forall x>1$ $exists y mid forall n geq y, F_n(x) notin mathbb{R}$.
Now, if we ever get a negative denominator, the end result will be nonreal. This is because addition, subtraction, and division between nonreals and nonzero reals will stay nonreal, and the square root of a nonreal will also be nonreal.
Now, when $n>2$ is odd, $sqrt{x^n}<x$, and thus $F_n(x) notin mathbb{R}$. Thus, we must concern ourselves with even $n$.
So, in $F_{2k}(x)$, consider:
$$ frac{cdots}{x – sqrt{ frac{x^{2k1}}{x+sqrt{x^{2k}}} }}$$
$$ x < sqrt{ frac{x^{2k1}}{x+x^k} } implies F_{2k}(x) notin mathbb{R} $$
Now, we will show that the “bottom” of $F_{2(k+1)}(x)$ is greater than the bottom of $F_{2k}(x)$, for $x>1$:
$$ sqrt{ frac{x^{2k1}}{x+sqrt{x^{2k}}} } < sqrt{ frac{x^{2k+1}}{x+sqrt{x^{2(k+1)}}} } $$
$$ frac{x^{2k1}}{x+x^k} < frac{x^{2k+1}}{x+x^{k+1}} $$
$$ x^{2k}+x^{3k} < x^{2k+2} + x^{3k+2} $$
The bottom inequality is true for $x>1$, thus by reversing the work, we prove the first line to be true. (we could also do this with partial derivatives, but that’s messier in my opinion.
Using this result, it follows that:
$$ x < sqrt{ frac{x^{2k1}}{x+x^k} } implies x < sqrt{ frac{x^{2k+1}}{x+x^{k+1}} } implies F_{2(k+1)}(x) notin mathbb{R} $$
Now, to prove the “bottom” will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $x > 1$. Since $x$ is nonnegative, the domain is $[0,1]$.
So, for any $x$ there’s an $y$ for which $F_n(x)$ is nonreal for finite $n$ greater than $y$. However, I’m not certain this is rigorously extends to the infinite case.
Extra notes: I’ve rigged up a computer program to calculate when $F_n(x)$ becomes nonreal. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+frac{1}{2^k}$.
$$smallbegin{array}{cc} k&0&1&2&3&4&5&6&7&8&9&10&11&12&13&14\hline n_{text{diverge}}&8&8&10&14&22&38&68&130&252&500&992&1978&3948&7890&15774end{array}$$
As you can see, as $x$ gets twice as close to $1$, it takes almost twice as many terms to go nonreal. I’ve tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $frac{1}{x1}$.
Proof attempt for monotonicity: (next improvement is to prove that $H'<1$)
Here I will attempt to prove the monotonicity of $F_infty$. First, let us introduce some definitions. $$F:=F_{infty},quad G:=sqrt{frac{x^2}{xsqrt{frac{x^3}{x+sqrt{frac{x^4}{x sqrt{ frac{x^5}{x+cdots}}}}}}}},quad H:=sqrt{frac{x^3}{x+sqrt{frac{x^4}{xsqrt{frac{x^5}{x+ sqrt{ frac{x^6}{x+cdots}}}}}}}}$$ Since $F=sqrt{dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F’=frac1{2F}cdotfrac{1(x+G)x(1+G’)}{(x+G)^2}>0impliedby GxG’>0$$ as $(x+G)^2$ and $F$ are clearly nonnegative.
Now this is implied by $$G’=frac1{2G}cdotleft(1+frac{H’x^2H}{(xH)^2}right)<frac Gx$$ and since $G=sqrt{dfrac{x^2}{xH}}$ (note that $H<x$), we get begin{align}2G^2>x+xfrac{H’x^2H}{(xH)^2}&impliedbyfrac{2x}{xH}>1+frac{H’x^2H}{(xH)^2}\&impliedby 2x^22Hx>x^22Hx+H^2+H’x^2H^2\&impliedby x^2>H’x^2impliedby H'<1end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H’$. Of course, the latter two are only close approximations to the real distribution of $H$.
$endgroup$

1$begingroup$
The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.
$endgroup$
– Zachary Hunter
Jan 5 at 18:18 
$begingroup$
Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.
$endgroup$
– Zachary Hunter
Jan 7 at 0:43 
1$begingroup$
Well done! It’s not too rigorous, but I can understand your method. I’ll make some edits later. I have made an attempt into showing its monotonicity, but I can’t get any farther than $H'<1$. Any help appreciated!
$endgroup$
– TheSimpliFire
2 days ago
add a comment 
Zachary’s and OP’s combined solutions
If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_infty$ at $x=1$ will give the maximum, which will be an infinite fraction:
$$F_infty (1) = sqrt{ frac{1}{1+sqrt{ frac{1}{1 sqrt{cdots} }} }} = sqrt{ frac{1}{1+sqrt{ frac{1}{1 F_infty (1) }} }}implies F_infty (1)^2 left(1+sqrt{ frac{1}{1 F_infty (1) }} right) = 1$$ so $$(F_infty(1)^21)^2=frac{F_infty(1)^4}{1F_infty(1)}implies F_infty(1)^52F_infty(1)^3+2F_infty(1)^2+F_infty(1)1=0.$$ Factoring out some roots, we get $$(F_infty(1)^2+F_infty(1)1)(F_infty(1)^3F_infty(1)^2+1)=0$$ It can be verified from WA, for example, that the only positive real solution is at $F_infty(1)=phi1$ derived from the first quadratic factor.
Proof attempt for the domain:
While I’m not sure how this holds up for $F_{infty}(x)$, we can show that $forall x>1$ $exists y mid forall n geq y, F_n(x) notin mathbb{R}$.
Now, if we ever get a negative denominator, the end result will be nonreal. This is because addition, subtraction, and division between nonreals and nonzero reals will stay nonreal, and the square root of a nonreal will also be nonreal.
Now, when $n>2$ is odd, $sqrt{x^n}<x$, and thus $F_n(x) notin mathbb{R}$. Thus, we must concern ourselves with even $n$.
So, in $F_{2k}(x)$, consider:
$$ frac{cdots}{x – sqrt{ frac{x^{2k1}}{x+sqrt{x^{2k}}} }}$$
$$ x < sqrt{ frac{x^{2k1}}{x+x^k} } implies F_{2k}(x) notin mathbb{R} $$
Now, we will show that the “bottom” of $F_{2(k+1)}(x)$ is greater than the bottom of $F_{2k}(x)$, for $x>1$:
$$ sqrt{ frac{x^{2k1}}{x+sqrt{x^{2k}}} } < sqrt{ frac{x^{2k+1}}{x+sqrt{x^{2(k+1)}}} } $$
$$ frac{x^{2k1}}{x+x^k} < frac{x^{2k+1}}{x+x^{k+1}} $$
$$ x^{2k}+x^{3k} < x^{2k+2} + x^{3k+2} $$
The bottom inequality is true for $x>1$, thus by reversing the work, we prove the first line to be true. (we could also do this with partial derivatives, but that’s messier in my opinion.
Using this result, it follows that:
$$ x < sqrt{ frac{x^{2k1}}{x+x^k} } implies x < sqrt{ frac{x^{2k+1}}{x+x^{k+1}} } implies F_{2(k+1)}(x) notin mathbb{R} $$
Now, to prove the “bottom” will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $x > 1$. Since $x$ is nonnegative, the domain is $[0,1]$.
So, for any $x$ there’s an $y$ for which $F_n(x)$ is nonreal for finite $n$ greater than $y$. However, I’m not certain this is rigorously extends to the infinite case.
Extra notes: I’ve rigged up a computer program to calculate when $F_n(x)$ becomes nonreal. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+frac{1}{2^k}$.
$$smallbegin{array}{cc} k&0&1&2&3&4&5&6&7&8&9&10&11&12&13&14\hline n_{text{diverge}}&8&8&10&14&22&38&68&130&252&500&992&1978&3948&7890&15774end{array}$$
As you can see, as $x$ gets twice as close to $1$, it takes almost twice as many terms to go nonreal. I’ve tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $frac{1}{x1}$.
Proof attempt for monotonicity: (next improvement is to prove that $H'<1$)
Here I will attempt to prove the monotonicity of $F_infty$. First, let us introduce some definitions. $$F:=F_{infty},quad G:=sqrt{frac{x^2}{xsqrt{frac{x^3}{x+sqrt{frac{x^4}{x sqrt{ frac{x^5}{x+cdots}}}}}}}},quad H:=sqrt{frac{x^3}{x+sqrt{frac{x^4}{xsqrt{frac{x^5}{x+ sqrt{ frac{x^6}{x+cdots}}}}}}}}$$ Since $F=sqrt{dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F’=frac1{2F}cdotfrac{1(x+G)x(1+G’)}{(x+G)^2}>0impliedby GxG’>0$$ as $(x+G)^2$ and $F$ are clearly nonnegative.
Now this is implied by $$G’=frac1{2G}cdotleft(1+frac{H’x^2H}{(xH)^2}right)<frac Gx$$ and since $G=sqrt{dfrac{x^2}{xH}}$ (note that $H<x$), we get begin{align}2G^2>x+xfrac{H’x^2H}{(xH)^2}&impliedbyfrac{2x}{xH}>1+frac{H’x^2H}{(xH)^2}\&impliedby 2x^22Hx>x^22Hx+H^2+H’x^2H^2\&impliedby x^2>H’x^2impliedby H'<1end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H’$. Of course, the latter two are only close approximations to the real distribution of $H$.
$endgroup$
Zachary’s and OP’s combined solutions
If we can prove it is monotonically increasing and has domain $[0,1]$, the limit is simple. Evaluating $F_infty$ at $x=1$ will give the maximum, which will be an infinite fraction:
$$F_infty (1) = sqrt{ frac{1}{1+sqrt{ frac{1}{1 sqrt{cdots} }} }} = sqrt{ frac{1}{1+sqrt{ frac{1}{1 F_infty (1) }} }}implies F_infty (1)^2 left(1+sqrt{ frac{1}{1 F_infty (1) }} right) = 1$$ so $$(F_infty(1)^21)^2=frac{F_infty(1)^4}{1F_infty(1)}implies F_infty(1)^52F_infty(1)^3+2F_infty(1)^2+F_infty(1)1=0.$$ Factoring out some roots, we get $$(F_infty(1)^2+F_infty(1)1)(F_infty(1)^3F_infty(1)^2+1)=0$$ It can be verified from WA, for example, that the only positive real solution is at $F_infty(1)=phi1$ derived from the first quadratic factor.
Proof attempt for the domain:
While I’m not sure how this holds up for $F_{infty}(x)$, we can show that $forall x>1$ $exists y mid forall n geq y, F_n(x) notin mathbb{R}$.
Now, if we ever get a negative denominator, the end result will be nonreal. This is because addition, subtraction, and division between nonreals and nonzero reals will stay nonreal, and the square root of a nonreal will also be nonreal.
Now, when $n>2$ is odd, $sqrt{x^n}<x$, and thus $F_n(x) notin mathbb{R}$. Thus, we must concern ourselves with even $n$.
So, in $F_{2k}(x)$, consider:
$$ frac{cdots}{x – sqrt{ frac{x^{2k1}}{x+sqrt{x^{2k}}} }}$$
$$ x < sqrt{ frac{x^{2k1}}{x+x^k} } implies F_{2k}(x) notin mathbb{R} $$
Now, we will show that the “bottom” of $F_{2(k+1)}(x)$ is greater than the bottom of $F_{2k}(x)$, for $x>1$:
$$ sqrt{ frac{x^{2k1}}{x+sqrt{x^{2k}}} } < sqrt{ frac{x^{2k+1}}{x+sqrt{x^{2(k+1)}}} } $$
$$ frac{x^{2k1}}{x+x^k} < frac{x^{2k+1}}{x+x^{k+1}} $$
$$ x^{2k}+x^{3k} < x^{2k+2} + x^{3k+2} $$
The bottom inequality is true for $x>1$, thus by reversing the work, we prove the first line to be true. (we could also do this with partial derivatives, but that’s messier in my opinion.
Using this result, it follows that:
$$ x < sqrt{ frac{x^{2k1}}{x+x^k} } implies x < sqrt{ frac{x^{2k+1}}{x+x^{k+1}} } implies F_{2(k+1)}(x) notin mathbb{R} $$
Now, to prove the “bottom” will always exceed $x$, simply note that as $n$ approaches infinity, the numerator grows faster than the denominator, and thus diverges when $x > 1$. Since $x$ is nonnegative, the domain is $[0,1]$.
So, for any $x$ there’s an $y$ for which $F_n(x)$ is nonreal for finite $n$ greater than $y$. However, I’m not certain this is rigorously extends to the infinite case.
Extra notes: I’ve rigged up a computer program to calculate when $F_n(x)$ becomes nonreal. For the following values of $k$, here is the smallest even $n$ where $F_n(x)$ diverges, where $x = 1+frac{1}{2^k}$.
$$smallbegin{array}{cc} k&0&1&2&3&4&5&6&7&8&9&10&11&12&13&14\hline n_{text{diverge}}&8&8&10&14&22&38&68&130&252&500&992&1978&3948&7890&15774end{array}$$
As you can see, as $x$ gets twice as close to $1$, it takes almost twice as many terms to go nonreal. I’ve tested this with different fractions, and the same pattern still holds, where $n$ is seemingly proportional to $frac{1}{x1}$.
Proof attempt for monotonicity: (next improvement is to prove that $H'<1$)
Here I will attempt to prove the monotonicity of $F_infty$. First, let us introduce some definitions. $$F:=F_{infty},quad G:=sqrt{frac{x^2}{xsqrt{frac{x^3}{x+sqrt{frac{x^4}{x sqrt{ frac{x^5}{x+cdots}}}}}}}},quad H:=sqrt{frac{x^3}{x+sqrt{frac{x^4}{xsqrt{frac{x^5}{x+ sqrt{ frac{x^6}{x+cdots}}}}}}}}$$ Since $F=sqrt{dfrac x{x+G}}$, for (increasing) monotonicity to occur, $$F’=frac1{2F}cdotfrac{1(x+G)x(1+G’)}{(x+G)^2}>0impliedby GxG’>0$$ as $(x+G)^2$ and $F$ are clearly nonnegative.
Now this is implied by $$G’=frac1{2G}cdotleft(1+frac{H’x^2H}{(xH)^2}right)<frac Gx$$ and since $G=sqrt{dfrac{x^2}{xH}}$ (note that $H<x$), we get begin{align}2G^2>x+xfrac{H’x^2H}{(xH)^2}&impliedbyfrac{2x}{xH}>1+frac{H’x^2H}{(xH)^2}\&impliedby 2x^22Hx>x^22Hx+H^2+H’x^2H^2\&impliedby x^2>H’x^2impliedby H'<1end{align} Unfortunately the fact that $H<x$ only cannot imply this; however, the following plot verifies the nice inequality. The dotted red line is the line $y=x$; the purple curve is $H$ (up to $x^{11}$) and the green curve is $H’$. Of course, the latter two are only close approximations to the real distribution of $H$.

1$begingroup$
The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.
$endgroup$
– Zachary Hunter
Jan 5 at 18:18 
$begingroup$
Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.
$endgroup$
– Zachary Hunter
Jan 7 at 0:43 
1$begingroup$
Well done! It’s not too rigorous, but I can understand your method. I’ll make some edits later. I have made an attempt into showing its monotonicity, but I can’t get any farther than $H'<1$. Any help appreciated!
$endgroup$
– TheSimpliFire
2 days ago
add a comment 

1$begingroup$
The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.
$endgroup$
– Zachary Hunter
Jan 5 at 18:18 
$begingroup$
Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.
$endgroup$
– Zachary Hunter
Jan 7 at 0:43 
1$begingroup$
Well done! It’s not too rigorous, but I can understand your method. I’ll make some edits later. I have made an attempt into showing its monotonicity, but I can’t get any farther than $H'<1$. Any help appreciated!
$endgroup$
– TheSimpliFire
2 days ago
The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.
$endgroup$
– Zachary Hunter
Jan 5 at 18:18
The domain limit seems fairly trivial. For the monotonic property, it might be helpful to notice that any $x<y$ both within the domain [0,1], $x=y^n$ for some $n>1$. However, at the moment, I cannot figure out exactly how to exploit the exponent yet.
$endgroup$
– Zachary Hunter
Jan 5 at 18:18
Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.
$endgroup$
– Zachary Hunter
Jan 7 at 0:43
Alrighty, added a legitimate proof, as long as my understanding of infinities is correct.
$endgroup$
– Zachary Hunter
Jan 7 at 0:43
Well done! It’s not too rigorous, but I can understand your method. I’ll make some edits later. I have made an attempt into showing its monotonicity, but I can’t get any farther than $H'<1$. Any help appreciated!
$endgroup$
– TheSimpliFire
2 days ago
Well done! It’s not too rigorous, but I can understand your method. I’ll make some edits later. I have made an attempt into showing its monotonicity, but I can’t get any farther than $H'<1$. Any help appreciated!
$endgroup$
– TheSimpliFire
2 days ago
add a comment 
Thanks for contributing an answer to Mathematics Stack Exchange!
 Please be sure to answer the question. Provide details and share your research!
But avoid …
 Asking for help, clarification, or responding to other answers.
 Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave(‘#loginlink’);
});
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin(‘.newpostlogin’, ‘https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062908%2fwhatisthemaximumvalueofthisnestedradical%23newanswer’, ‘question_page’);
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave(‘#loginlink’);
});
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave(‘#loginlink’);
});
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave(‘#loginlink’);
});
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
How exactly does the expression for $F_k$ look? I can’t actually work out easily how it terminates at the k^th step.
$endgroup$
– T_M
Jan 5 at 17:20