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?
 A: 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
A: Assuming both nested square roots are well-defined, we have $a=\sqrt{7b}$ and $b=\sqrt{2a}$, from which
$ab=\sqrt{14 ab}$ and $ab=\color{blue}{14}$.
A: 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: $$a=\sqrt{7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}}$$
$$a^2=7\sqrt{2\sqrt{7\sqrt{2\sqrt{...}}}}$$
$$a^4=98\sqrt{7\sqrt{2\sqrt{...}}}$$
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 $ab=xy$.
