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
  • 2
    $\begingroup$ 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 $ab=\color{blue}{14}$.

  • 1
    $\begingroup$ i would add something like $ab \neq 0$ $\endgroup$
    – user1
    Aug 19, 2017 at 23:15
  • 3
    $\begingroup$ @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
  • 1
    $\begingroup$ 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 $ab=xy$.

  • 8
    $\begingroup$ 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
  • 2
    $\begingroup$ IMHO unless you prove that the limits exist, the answer is not useful. $\endgroup$ Aug 19, 2017 at 16:59
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.