Would like to derive an analytic solution to the following two infinite products: $$P_1=\frac{4}{5}\frac{8}{7}\frac{10}{11}\frac{14}{13}\frac{16}{17}...$$ $$=\prod_{n=1}^\infty \frac{3+6n+(-1)^n} {3+6n-(-1)^n}$$ and $$P_2=\frac{8}{7}\frac{13}{14}\frac{20}{19}\frac{25}{26}\frac{32}{31}...$$ $$=\prod_{n=1}^\infty\frac{3+12n-(-1)^n} {3+12n+(-1)^n}$$
These are similar to the Willis product of $$P_W=\frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}\frac{6}{5}...$$ $$=\prod_{n=0}^\infty\frac{3+2n+(-1)^n} {3+2n-(-1)^n}$$ featured on the latest episode of 3Blue1Brown which is known to converge to $\frac{\pi}{2}$. Doing these numerical by hand, $P_1$ converges to around 0.87 and $P_2$ to around 1.09. Are the real convergent values irrational (thinking yes) and if so, are they multiples of powers of $\pi$? Also, are the solutions extensible/describable via parametrization to a series?
These products can be computed through standard Weierstrass products. For instance
$$ \prod_{n=1}^{2N}\frac{3+6n+(-1)^n}{3+6n-(-1)^n} = \prod_{n=1}^{N}\frac{4+12n}{2+12n}\cdot\frac{-4+12n}{-2+12n}=\frac{\sqrt{3}\,\Gamma\left(N+\tfrac{2}{3}\right)\,\Gamma\left(N+\tfrac{4}{3}\right)}{2\,\Gamma\left(N+\tfrac{5}{6}\right)\,\Gamma\left(N+\tfrac{7}{6}\right)} $$ converges to $\frac{\sqrt{3}}{2}$ as $N\to +\infty$, and $$ \prod_{n=1}^{2N}\frac{3+12n+(-1)^n}{3+12n-(-1)^n} = \prod_{n=1}^{N}\frac{4+24n}{2+24n}\cdot\frac{-10+24n}{-8+24n}$$ converges to $\frac{\sqrt{\pi}\,\Gamma\left(\frac{2}{3}\right)}{2^{1/6}\,\Gamma\left(\frac{7}{12}\right)^2}$.
TLDR: you only need the Weierstrass product for the $\Gamma$ function to manage those products.