I am wondering whether there exists an easy way to evaluate the following infinite product :
$\sqrt{\frac{1}{3}}.\sqrt{\frac{1}{3}+\frac{1}{3}.\sqrt{\frac{1}{3}}}.\sqrt{\frac{1}{3}+\frac{1}{3}.\sqrt{\frac{1}{3}+\frac{1}{3}.\sqrt{\frac{1}{3}}}}...$
or even better (assuming convergence) of :
$\sqrt{\frac{1}{a}}.\sqrt{\frac{1}{a}+\frac{1}{a}.\sqrt{\frac{1}{a}}}.\sqrt{\frac{1}{a}+\frac{1}{a}.\sqrt{\frac{1}{a}+\frac{1}{a}.\sqrt{\frac{1}{a}}}}...$
with $a>0.$
This is a variant of Viète's formula :
$V_{\infty}=\sqrt{\frac{1}{2}}.\sqrt{\frac{1}{2}+\frac{1}{2}.\sqrt{\frac{1}{2}}}.\sqrt{\frac{1}{2}+\frac{1}{2}.\sqrt{\frac{1}{2}+\frac{1}{2}.\sqrt{\frac{1}{2}}}}...=\frac{2}{\pi}$
which is much easier to prove since we can define :
$u_{n+1}=\sqrt{\frac{1+u_n}{2}}$ and realise that for $|u_0|<1$ we can also define $cos(\theta)=u_0$. Viète's formula is just $\prod_{n=1}^{\infty}u_n$, which can be easily evaluated using trigonometric identities, especially $cos(2\theta)=2cos²(\theta)-1$ which yields : $cos(\frac{\theta}{2})=\sqrt{\frac{1+cos(\theta)}{2}}$
So $\prod_{n=1}^{\infty}u_n=\prod_{n=1}^{\infty}cos(\frac{\theta}{2^n})$ Then using $sin(2\theta)=2sin(\theta)cos(\theta)$ mass cancellation occurs and we find $\prod_{n=1}^{\infty}u_n=\frac{sin(\theta)}{\theta}.$ Viète's formula corresponds to $u_0=0$, so $\theta=\frac{\pi}{2}$. By replacing with this value, we find Viète's value.
Unfortunately, this strategy doesn't seem to work when the 2's are replaced by something else (like 3 for example) (or at least, I can't find any tractable trigonometric formulas to help) Any idea ? Thank you !
(P.S. : I am very aware that Viète did not use this strategy to find this formula. He used geometry arguments I do not know, so perhaps there is a way to generalize using geometry, although I'd rather have a proof using only analysis...)
Where does this problem come from? Anyway, for $a>0$, let $u_0^{(a)} = 0$ and $u_{n+1}^{(a)} = \sqrt{\frac{1+u_n^{(a)}}{a}} = \sqrt{\frac{2}{a}}\sqrt{\frac{1+u_n^{(a)}}{2}}$. We wish to determine $\prod_{n=1}^{\infty}u_n^{(a)}$. Repeating the trigonometric substitutions in your post, we find $$\prod_{n=1}^{\infty}u_n^{(a)} = \frac{2}{\pi}\cdot \lim_{n \to \infty} \left(\sqrt{\frac{2}{a}} \right)^n $$ For $0<a<2$ this diverges, for $a=2$ it equals $2/\pi$ and for $a>2$ it equals $0$.