<s>[INST] a conjectured continued fraction for $\tan\left(\frac{z\pi}{4z+2n}\right)$ 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.
 [/INST] The proposed continued fraction
\begin{equation}
\displaystyle\tan\left(\frac{z\pi}{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{equation} 
can be written as
\begin{equation}
\displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)=\cfrac{2z/\left( 2z+n \right)}{1+\cfrac{(n)/\left( 2z+n \right)\cdot(4z+n)/\left( 2z+n \right)} {3+\cfrac{(2z+2n)/\left( 2z+n \right)\cdot(6z+2n)/\left( 2z+n \right)}{5+\cfrac{(4z+3n)/\left( 2z+n \right)\cdot(8z+3n)/\left( 2z+n \right)}{7+\ddots}}}}
\end{equation} 
Denoting $u=\cfrac{z}{4z+2n}$, the factors of the numerators are
\begin{equation}
\frac{n}{2z+n}=1-4u\,;\quad\frac{4z+n}{2z+n}=1+4u\,;\quad\frac{2z+2n}{2z+n}=2-4u\,;\quad\frac{6z+2n}{2z+n}=2+4u\,;\cdots
\end{equation} 
Then, the fraction can be simplified as
\begin{equation}
\displaystyle\tan\left(\pi u\right)=\cfrac{4u}{1+\cfrac{\cfrac{1-16u^2}{1\cdot3}} {1+\cfrac{\cfrac{4-16u^2}{3\cdot5}}{1+\cfrac{\cfrac{9-16u^2}{5\cdot7}}{1+\ddots}}}}
\end{equation} 
It is thus a special case of the continued fraction found in this answer:
\begin{equation}
\tan\left(\alpha\tan^{-1}z\right)=\cfrac{\alpha z}{1+\cfrac{\frac{(1^2-\alpha^2)z^2}{1\cdot 3}} {1+\cfrac{\frac{(2^2-\alpha^2)z^2}{3\cdot 5}}{1+\cfrac{\frac{(3^2-\alpha^2)z^2}{5\cdot 7}}{1+\ddots}}}}
\end{equation} 
here $z=1$ and $\alpha=4u$. The brilliant proof is based on a continued fraction due to Nörlund.
</s>