TITLE: a conjectured continued fraction for $\tan\left(\frac{z\pi}{4z+2n}\right)$
QUESTION [45 upvotes]: Given a complex number $\begin{aligned}\frac{z}{n}=x+iy\end{aligned}$ and a gamma function $\Gamma(z)$ with $x\gt0$, it is conjectured that the following continued fraction for $\displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)$ is true
$$\begin{split}\displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)&=\frac{\displaystyle\Gamma\left(\frac{z+n}{4z+2n}\right)\Gamma\left(\frac{3z+n}{4z+2n}\right)}{\displaystyle\Gamma\left(\frac{z}{4z+2n}\right)\Gamma\left(\frac{3z+2n}{4z+2n}\right)}\\&=\cfrac{2z}{2z+n+\cfrac{(n)(4z+n)} {3(2z+n)+\cfrac{(2z+2n)(6z+2n)}{5(2z+n)+\cfrac{(4z+3n)(8z+3n)}{7(2z+n)+\ddots}}}}\end{split}$$
Corollaries:
By taking the limit(which follows after abel's theorem) 
$$
\begin{aligned}\lim_{z\to0}\frac{\displaystyle\tan\left(\frac{z\pi}{4z+2}\right)}{2z}=\frac{\pi}{4}\end{aligned},
$$ 
we recover the well known continued fraction for $\pi$
$$\begin{aligned}\cfrac{4}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\ddots}}}}=\pi\end{aligned}$$
If we let $z=1$ and $n=2$,then we have the square root of $2$ $$\begin{aligned}{1+\cfrac{1}{2+\cfrac{1} {2+\cfrac{1}{2+\cfrac{1}{2+\ddots}}}}}=\sqrt{2}\end{aligned}$$
Q: How do we prove rigorously that the conjectured continued fraction is true and converges for all complex numbers $z$ with $x\gt0$?
Update:I initially defined the continued fraction $\displaystyle\tan\left(\frac{z\pi}{4z+2}\right)$ for only natural numbers,but as a matter of fact it holds for all complex numbers $z$ with real part greater than zero.Moreover,this continued fraction is a special case of the general continued fraction found in this post.

REPLY [5 votes]: The ratio
$$\tan\dfrac{\pi z}{4z+2n} 
= \dfrac{\Gamma\left(\dfrac{z+n}{4z+2n}\right)\Gamma\left(\dfrac{3z+n}{4z+2n}\right)}{\Gamma\left(\dfrac{z}{4z+2n}\right)\Gamma\left(\dfrac{3z+2n}{4z+2n}\right)}\hspace{100mu}\tag1$$
can be obtained, applying "real" identity

$$\sin\pi x = \dfrac\pi{\Gamma(x)\Gamma(1-x)}\hspace{100mu}\tag2$$

to the expression
$$\tan\dfrac\pi2\dfrac z{2z+n} 
= \dfrac{\sin\pi\dfrac z{4z+2n}}{\sin\pi\dfrac{z+n}{4z+2n}},$$
so it looks nice and quite correct.
Continued fraction can be obtained, using known continued fraction of the tangent function in the form of
$$\tan \dfrac{\pi x}4 = \cfrac x{1+\operatorname{
\Large K}\hspace{-27mu}\phantom{\Big|}_{k=1}^{\large ^{\,\infty}}\cfrac{(2k-1)^2-x^2}2}\hspace{100mu}\tag3$$
with 
$$x=\dfrac{2z}{2z+n}.$$