I found a question that asked to find the limiting value of $$10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}$$If you make the substitution $x=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}$ it simplifies to $x=10\sqrt{x}$ which has solutions $x=0,100$. I don't understand how $x=0$ is a possible solution, I know that squaring equations can introduce new, invalid solutions to equations and so you should check the solutions in the original (unsquared) equation, but doing that here doesn't lead to any non-real solutions or contradictions. I was wondering if anyone knows how $x=0$ turns out as a valid solution, is there an algebaric or geometric interpretation? Or is it just a "special case" equation?

A similar question says to find the limiting value of $\sqrt{6+5\sqrt{6+5\sqrt{6+5\sqrt{...}}}}$, and making a similar substituion for $x$ leads to $$x=\sqrt{6+5x}$$ $$x^2=6+5x$$ which has solutions $x=-1,6$. In this case though, you could substitute $x=-1$ into the first equation, leading to the contradiction $-1=1$ so you could satisfactorily disclude it.

Is there any similar reasoning for the first question? I know this might be a stupid question but I'm genuinely curious :)

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    $\begingroup$ The problem is that you question is not well formulated, since it does not say how those expressions end up (or should I say start out) at the innermost level. Should they end with $\cdots\sqrt0$, then they will all evaluate to $0$, which gives a perfectly converging (in fact constant) sequence. This is probably not the intended interpretation, but infinite expressions simply don't make sense; one needs a sequence of definite values to take a limit. $\endgroup$ Oct 2, 2014 at 8:32
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    $\begingroup$ You argued $$x=10\sqrt{\mathstrut\ldots}\ \Rightarrow\ \ldots \Rightarrow\ x\in\{0,100\}\ ,$$ but from this you cannot conclude that $0$ is a solution of the original problem. $\endgroup$ Oct 2, 2014 at 9:26
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    $\begingroup$ This is not a stupid question at all. I'm a math grad student right now and don't know either. $\endgroup$
    – BCLC
    Dec 6, 2014 at 12:44
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    $\begingroup$ See the other answers on proving your answer. You merely guess x = 100. You have to prove it. In fact, how do you know such a limit exists? Sure x would exist if process did not go on infinitely but what about when it does? $\endgroup$
    – BCLC
    Dec 6, 2014 at 12:47
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    $\begingroup$ Possible duplicate of $\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation $\endgroup$ Feb 28, 2017 at 16:43

8 Answers 8


Denote the given problem as $x$, then \begin{align} x&=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{\cdots}}}}}\\ &=10\cdot10^{\large\frac{1}{2}}\cdot10^{\large\frac{1}{4}}\cdot10^{\large\frac{1}{8}}\cdot10^{\large\frac{1}{16}}\cdots\\ &=10^{\large1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots}\\ &=10^{\large y} \end{align} where $y$ is an infinite geometric series in which its value is \begin{align} y &=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots\\ &=\frac{1}{1-\frac{1}{2}}\\ &=2 \end{align} Therefore \begin{equation} x=10^{\large 2}=100 \end{equation}

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    $\begingroup$ Nice And interesant resolution! $\endgroup$ Nov 3, 2014 at 17:12
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    $\begingroup$ The English word for 'interesante' is 'interesting' ;) $\endgroup$ Dec 6, 2014 at 12:05
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    $\begingroup$ Inventive point of view. We are trained to see the recursive structure immediately, but you can see directly the complete solution! :-) +1 $\endgroup$
    – epi163sqrt
    Dec 15, 2014 at 15:23
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    $\begingroup$ Never seen this approach before. Beautiful and elegant! :) $\endgroup$
    – MonK
    Jun 3, 2015 at 10:25
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    $\begingroup$ Haha, is it too late to praise you again? $\endgroup$ Jan 5, 2017 at 21:56

If $$x=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}$$ then it is true that $x=10\sqrt x$.

It is not true that if $x=10\sqrt x$, then $$x=10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}.$$

$0$ is indeed one solution, but it is not a valid one.

  • $\begingroup$ Well the OP asked why the solution is not the valid one,I mean there is no dividing by zero,or logarithm with $x<0$ or negative values under sqrt etc. $\endgroup$
    – kingW3
    Oct 2, 2014 at 9:54
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    $\begingroup$ @kingW3: I think that this answer (and Christian Blatter's comment under the original post) exactly address the root cause of this: implication is a one-way street, equivalence (or "if and only if") is two-way . The concerns you listed are school level examples of why we sometimes get one way traffic only. Trying to memorize such rules is wasted effort, because, as witnessed here, it is a poor substitute to understanding the logic of the situation. $\endgroup$ Oct 6, 2014 at 7:58
  • $\begingroup$ Great answer! +1 However maybe emphasise more on the not $\iff$ part of the answer because people still are not getting it. $\endgroup$ Oct 6, 2014 at 17:29
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    $\begingroup$ No explanation about why $0$ is not a valid solution? $\endgroup$
    – Did
    Aug 21, 2017 at 17:09
  • $\begingroup$ @Did In some way, the questioner only asked "how does $0$ appear as a candidate solution", and I answered only that question as shortly as possible.I think that other answers covered other parts of the question very well. $\endgroup$
    – 5xum
    Aug 21, 2017 at 20:56

Here the question is (apparently) about the sequence defined recursively by $$ \begin{aligned} a_1&=10,\\ a_{n+1}&=10\sqrt{a_n}\quad \text{for all positive integers $n$.} \end{aligned} $$ It often happens that students embark on a quest of finding the value of the limit of a sequence before ascertaining that what they try to calculate actually exists. Here we should first satisfy ourselves that $\lim_{n\to\infty}a_n$ actually exists as a real number. A convenient tool for this is the theorem telling us that a bounded increasing sequence converges towards some limit $A$. Here it is easy to prove by induction that $a_{n+1}>a_n$ for all $n$, and also that $a_n<100$ for all $n$. A part of that theorem then says that $a_1\le A\le 100$. After all, all the members of the sequence are in the closed interval $[10,100]$ so their limit cannot be outside this interval either. The recurrence relation (and continuity of the square root function in this interval) then give us the equation $A=10\sqrt{A}$ allowing us to deduce that either $A=0$ or $A=100$. The former solution, however, does not belong to this interval and can be discarded as an alternative.

Your equation having another solution is just a coincidence. In this case as well as with the recurrence formula $a_{n+1}=\sqrt{6+5a_n}$.

Compare with the following other common misuse of limits: $$ S=1-1+1-1+1-1+\cdots $$ Some are want for rewriting this as $$ S=1-(1-1+1-1+1-\cdots)=1-S, $$ solving the equation $S=1-S$, and concluding that $S=1/2$.

Where's the mistake? It is in the first line, where the student treats the sum of this series as if it has a value, i.e. as if it converges, by the act of calling it $S$. The moral: Beware of naming things that don't necessarily exist :-)

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    $\begingroup$ Only answer, which says, why 0 is not valid solution. $\endgroup$
    – Somnium
    Jun 17, 2016 at 9:09

Alternatively, let $a_1,\,a_2,\,a_3,\,\cdots,\,a_n$ be the following sequence $$10\sqrt{10},\,10\sqrt{10\sqrt{10}},\,10\sqrt{10\sqrt{10\sqrt{10}}},\,\cdots,10\underbrace{\sqrt{10\sqrt{10\sqrt{10\sqrt{\cdots\sqrt{10}}}}}}_{\large n\,\text{times}}$$ respectively.

Notice that $$\large a_n=10^{\Large 2-2^{-n}}$$ Hence $$10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{\cdots}}}}}=\large\lim_{n\to\infty}\, a_n=\lim_{n\to\infty}\,10^{\Large 2-2^{-n}}=\bbox[3pt,border:3px #FF69B4 solid]{\color{red}{100}}$$


Consider the sequence $a_{n+1}= 10\sqrt{a_n}$ with $a_0 \ge 0$. The expression you have could be considered the limit as $n \to \infty$, if it exists.

Further, $a_{n+1} = \sqrt{100 \cdot a_n}$ gives that $a_{n+1}$ is between $100$ and $a_n$. So if you start with $a_0 > 0$, you have a nice bounded monotone sequence, which converges to a limit and the limit will of course be $100$.

OTOH if you start with $a_0 = 0$, clearly the limit exists still, and you have the limit as $0$.

As your expression is ambiguous about what $a_0$ could be, you get both solutions.


$$x= 10\sqrt{10\sqrt{10\sqrt{10\sqrt{\cdots}}}} \Rightarrow \left(\frac{x}{10}\right)^2 = x$$ $$\Rightarrow x^2 -100x = 0 \Rightarrow x(x-100) = 0$$ Since $x$ is positive (why?), we conclude that $x=100$.

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    $\begingroup$ "$\Rightarrow x(x-100) = 0\ x=100$" I am following you (taking for granted that the limit $x$ exists) until "$x(x-100) = 0$". But then, how do you deduce "$x=100$"? $\endgroup$
    – Did
    Aug 22, 2017 at 6:51
  • $\begingroup$ Thanks @Did for your observation. $\endgroup$ Sep 3, 2017 at 2:28

have you noticed that $10$ is no special value. In this replace $10$ by $a$ i.e. any real number, equation will be still $x=a\sqrt{x}$, and $0$ will be a solution, so this can be regarded as a defect in such equations and thus always ignored, I don't think there is any geometrical or algebraic significance of this.

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    $\begingroup$ I believe the OP is well aware that $a = 10$ is just a special case. He is asking what exactly is the cause of defect in such equations. $\endgroup$
    – taninamdar
    Oct 2, 2014 at 8:39
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    $\begingroup$ The reason why $0$ is not a solution (rather than being "always ignored") is definitely not the "defect in such equations " that $x=0$ solves simultaneously every equation $x=a\sqrt{x}$. Thus, this (upvoted) answer seems to miss the point. $\endgroup$
    – Did
    Aug 22, 2017 at 6:55

We have:
$a_{n}=10\underbrace{\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...\sqrt{10}}}}}}}_{n\text{ times }}=10\times(\dfrac{2\sqrt{5}}{\sqrt{2}})^{\frac{2^{0}+2^{1}+2^{2}+2^{3}+.....+2^{n-1}}{2^{n-1}}}$
$a_{n}=10\underbrace{\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...\sqrt{10}}}}}}}_{n\text{ times }}=10\times(\dfrac{2\sqrt{5}}{\sqrt{2}})^{\frac{\displaystyle\sum_{k=0}^{n-1}2^{k}}{2^{n-1}}}$
$a_{n}=10\underbrace{\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...\sqrt{10}}}}}}}_{n\text{ times }}=10\times(\dfrac{2\sqrt{5}}{\sqrt{2}})^{2\times\dfrac{\displaystyle\sum_{k=0}^{n-1}2^{k}}{2^{n}}}$\ $a_{n}=10\underbrace{\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...\sqrt{10}}}}}}}_{n\text{ times }}=10\times(\dfrac{2\sqrt{5}}{\sqrt{2}})^{2\times\frac{\frac{2^{n-1+1}-1}{2-1}}{2^{n}}}$
$a_{n}=10\underbrace{\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...\sqrt{10}}}}}}}_{n\text{ times }}=10\times(\dfrac{2\sqrt{5}}{\sqrt{2}})^{2\times\frac{2^{n}-1}{2^{n}}}$
$a_{n}=10\underbrace{\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...\sqrt{10}}}}}}}_{n\text{ times }}=10\times(\dfrac{2\sqrt{5}}{\sqrt{2}})^{2\times(1-\frac{1}{2^{n}})}$
$10\sqrt{10\sqrt{10\sqrt{10\sqrt{10\sqrt{...}}}}}=\displaystyle \lim_{n\rightarrow\infty} a_{n}=\displaystyle \lim_{n\rightarrow\infty} 10\times(\dfrac{2\sqrt{5}}{\sqrt{2}})^{2\times(1-\frac{1}{2^{n}})}=10\times(\dfrac{2\sqrt{5}}{\sqrt{2}})^{2}=10*10=100$\


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