Given $a$ and $b$ calculate $ab$ $$a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}$$ $$b=\sqrt{2\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}}$$

I simplified the terms and further obtained that $ab$ is equal to: $$ab=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}...}\cdot7^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...}$$

How can I get a finite value?

  • $\begingroup$ See math.stackexchange.com/questions/589288/… $\endgroup$ Aug 19, 2017 at 16:15
  • $\begingroup$ @JoshuaSalazar This can be generalized for any $x,y$ inside the radicals, not just $2$ and $7$. See my answer. $\endgroup$ Aug 19, 2017 at 16:28
  • $\begingroup$ See also this for more discussion about the convergence issues. $\endgroup$ Aug 19, 2017 at 17:32
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    $\begingroup$ How is @Famkes second answer not obvious? What keeps us from immediately substituting $1$ for $\frac{1}{2}+\frac{1}{4}+...$ ? $\endgroup$ Aug 19, 2017 at 18:19

4 Answers 4


Assuming both nested square roots are well-defined, we have $a=\sqrt{7b}$ and $b=\sqrt{2a}$, from which $ab=\sqrt{14 ab}$ and $ab=\color{blue}{14}$.

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    $\begingroup$ i would add something like $ab \neq 0$ $\endgroup$
    – user1
    Aug 19, 2017 at 23:15
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    $\begingroup$ @Ben: I think it is trivial that if $a$ and $b$ are well defined they are positive, so $ab$ is positive as well. $\endgroup$ Aug 19, 2017 at 23:16
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    $\begingroup$ It surely isn't hard to show that $a,b>0$, but by definition a squareroot can be $0$. $\endgroup$
    – user1
    Aug 20, 2017 at 11:41

First answer:

Notice that:

$$\color{Blue}{a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}} \ \ \ ;$$ $$\color{Red}{b=}\sqrt{\color{Red}{2}\color{Blue}{\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}}} \ \ \ ;$$

which implies that:

$$ \color{Red}{b}=\sqrt{\color{Red}{2}\color{Blue}{a}} \ \ \ ; $$

similarly we have:

$$a=\sqrt{7b} \ \ \ . $$

So we must have:

$$a= \sqrt{7b}= \sqrt{7\sqrt{2a}} = \sqrt[4]{98a} \Longrightarrow a^4=98a \Longrightarrow a^4-98a=0 ; $$

but notice that the equation $x(x^3-98)$ has only two real solutions; $0$ and $\sqrt[3]{98}$.
So we can conclude that $a=\sqrt[3]{98}$.

Also we must have:

$$b= \sqrt{2a}= \sqrt{2\sqrt{7b}} = \sqrt[4]{28b} \Longrightarrow b^4=28b \Longrightarrow b^4-28b=0 ; $$

but notice that the equation $x(x^3-28)$ has only two real solutions; $0$ and $\sqrt[3]{28}$.
So we can conclude that $b=\sqrt[3]{28}$.

So we have: $ab=\sqrt[3]{98}\sqrt[3]{28}=\sqrt[3]{2^3.7^3}=\color{Green}{14}.$

Second answer: Notice that

$$ \color{Green}{\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + ... = 1 } ; $$

so we can conclude that $ab=2^1.7^1=\color{Green}{14}$

  • $\begingroup$ @ Jyrki Lahtonen you are right. It need to attend some analytic conditions; which I have been forgot. $\endgroup$
    – Davood
    Aug 19, 2017 at 16:48
  • $\begingroup$ a,b = 0 is clearly an extraneous root that is incompatible with how they are defined $\endgroup$
    – smci
    Aug 20, 2017 at 21:39

$$a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}$$ $$a^2=7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}$$ $$a^4=98\sqrt{7\sqrt{2\sqrt{...}}}$$ so $$a^4=98a$$ and, assuming $a$ is nonzero, $$a=\sqrt[3]{98}$$

$$b=\sqrt{2\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}}$$ $$b^2=2\sqrt{7\sqrt{2\sqrt{7\sqrt{...}}}}$$ $$b^4=28\sqrt{2\sqrt{7\sqrt{...}}}$$ $$b^4=28b$$ and, assuming $b$ is nonzero, $$b=\sqrt[3]{28}$$ so $$ab=\sqrt[3]{2744}=14$$

Additionally, it's not hard to prove that if $$a=\sqrt{x\sqrt{y\sqrt{x\sqrt{y\sqrt{...}}}}}$$ and $$b=\sqrt{y\sqrt{x\sqrt{y\sqrt{x\sqrt{...}}}}}$$ then $ab=xy$.

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    $\begingroup$ Are you also one of those people who think that $1-1+1-1+1-\cdots=1/2$ because $S=1-1+1-1+\cdots$ satisfies $$1-S=1-(1-1+1-1+\cdots)=1-1+1-1+1-1\cdots=S?$$ $\endgroup$ Aug 19, 2017 at 16:51
  • $\begingroup$ @JyrkiLahtonen I don't see how that is analogous to this problem. The limit as $n$ goes to infinity of $$\sum_{x=0}^n (-1)^x$$ does not exist because of fluctuation, whereas this sequence does not fluctuate. I don't think my question deserves to be down voted because of my recursive definition - it is appropriate in this problem, but not in your example. $\endgroup$ Aug 19, 2017 at 16:57
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    $\begingroup$ IMHO unless you prove that the limits exist, the answer is not useful. $\endgroup$ Aug 19, 2017 at 16:59
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    $\begingroup$ What @JyrkiLahtonen means is that you need to ensure the given infinitely nested radical converges in some sense. This usually involves something along the lines$$a_1=\sqrt{b_1}\\a_2=\sqrt{b_1\sqrt{b_2}}\\a_3=\sqrt{b_1\sqrt{b_2\sqrt{b_3}}}\\\vdots\\\sqrt{b_1\sqrt{b_2\sqrt{b_3\sqrt{\dots}}}}\equiv\lim_{n\to\infty}a_n$$If this limit does not exist, we would usually say the nested radical does not converge, though your solution ignores this. $\endgroup$ Aug 19, 2017 at 20:20
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    $\begingroup$ Of course, proving the convergence is not too hard to do. Indeed, the related sequences here are clearly monotonically increasing, and by induction, bounded above by your claimed values, and thus convergent. $\endgroup$ Aug 19, 2017 at 20:23

This seems easier than the half page proofs people are providing

$$a = \sqrt{7 b}$$

$$b = \sqrt{2 a}$$

$$a^2 = 7 b$$

$$b^2 = 2 a$$

$$a^2 b^2 = 14 a b$$

$$a b = 14$$

Unless I am missing something, a > 0 and b >0 we already know


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