Solving an infinite product of consecutive square roots
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?
Assuming both nested square roots are well-defined, we have $a=\sqrt{7b}$ and $b=\sqrt{2a}$, from which $ab=\sqrt{14 ab}$ and $ab=\color{blue}{14}$.
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 $ab=2^1.7^1=\color{Green}{14}$
$$a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}$$ $$a^2=7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}$$ $$a^4=98\sqrt{7\sqrt{2\sqrt{...}}}$$ 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 $ab=xy$.
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