Find the value of $P=\frac{\sqrt{6}}{3}\cdot\frac{\sqrt{6+\sqrt{6}}}{3}\cdot\frac{\sqrt{6+\sqrt{6+\sqrt{6}}}}{3}\cdots$. I made this problem inspired by Viete's formula.
I know for sure that it converges because $\sqrt{6+\sqrt{6+\sqrt{6+...}}}$ approaches the value of 3. I have also tested numerically that it approaches $\approx .7862$. I have tried repeatedly using the cosine half angle identity with $\cos\theta=\frac{\sqrt{6}}{3}$, but this doesn't work very nicely with the formula $\cos \frac{\theta}{2}=\sqrt{\frac{1+\cos\theta}{2}}$. I don't know of any other methods to deal with infinite products, but maybe there exists an elegant geometric solution like the 2n-gon proof of Viete's formula.
Also, for a more generalized problem, what is the value of $P_m=\prod_{i=1}^\infty \frac{a_i}{m}$, where $a_i=\sqrt{m(m-1)+a_{i-1}}$ and $a_0=0$. This is the same as the above problem for $m=3$ and $P_2=\frac{2}{\pi}$ by Viete's formula.
