TITLE: Solving an infinite product of consecutive square roots
QUESTION [13 upvotes]: 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?

REPLY [6 votes]: 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