Is there a closed form for $3\cdot\frac{3}{\sqrt{6}}\cdot\frac{3}{\sqrt{6+\sqrt{6}}}\cdot\frac{3}{\sqrt{6+\sqrt{6+\sqrt{6}}}}\cdots$?

Inspired by Vieta's Formula for $\pi$, $$\pi=2\cdot\frac{2}{\sqrt{2}}\cdot\frac{2}{\sqrt{2+\sqrt{2}}}\cdot\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\cdots$$ I became interested in a more generalized case for Vieta's Formula.

For $m=\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+\cdots}}}}$, what is the closed form of $$m\cdot\frac{m}{\sqrt{a}}\cdot\frac{m}{\sqrt{a+\sqrt{a}}}\cdot\frac{m}{\sqrt{a+\sqrt{a+\sqrt{a}}}}\cdots$$?

To find the value of $m$, we can solve the following equation $$m=\sqrt{a+m}$$, which gives $$m=\frac{1+\sqrt{1+4a}}{2}$$

One example is that $$3=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}}$$ And I want to find the exact value of $$3\cdot\frac{3}{\sqrt{6}}\cdot\frac{3}{\sqrt{6+\sqrt{6}}}\cdot\frac{3}{\sqrt{6+\sqrt{6+\sqrt{6}}}}\cdots\approx3.815$$ but to no avail. Could anyone provide any insight on the possible closed form for this infinite product?

Solution 1:

Just a comment, as the comment section is really getting long, you may really want to know the connection why it is $2$ in the radical that gives this amazing result. It is due to it's connection to trigonometric functions. More specifically, a quick research into this can give that a result, (link) $$\frac{\sin x}{x}=\cos \frac{x}{2}\cdot\cos \frac{x}{4}\cdot\cos \frac{x}{8}...$$ And also the fact that $$\cos\frac{x}{2^n}=\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+...(n-1) times}}}$$ or more specifically speaking; $$A=\frac{\sin A}{\cos\frac{A}{2}\cdot \cos\frac{A}{4}\cdot \cos\frac{A}{8}...}$$ So a generalized version of this would be doubtful. And also one can obviously make a conjecture that that your stated problem will result in a transendental number which couldn't be expressed in terms of closed form algebraic numbers or functions.

A very good paper, if you're really interested in square roots will be this, by Dixon J. Jones