a conjectured continued-fraction for $\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)$ that leads to a new limit for $\pi$ 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 $\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)$ is true
$$\begin{aligned}\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)=\frac{\displaystyle\Gamma\left(\frac{z}{4z+2n}\right)\Gamma\left(\frac{3z+2n}{4z+2n}\right)}{\displaystyle\Gamma\left(\frac{z+n}{4z+2n}\right)\Gamma\left(\frac{3z+n}{4z+2n}\right)}=\cfrac{2z+2n}{2z+n+\cfrac{(0z-n)(4z+3n)} {3(2z+n)+\cfrac{(2z+0n)(6z+4n)}{5(2z+n)+\cfrac{(4z+n)(8z+5n)}{7(2z+n)+\cfrac{(6z+2n)(10z+6n)}{9(2z+n)+\ddots}}}}}\end{aligned}$$
Or in gauss's notation $$\begin{aligned}\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)=-\frac{1}{2z+2n}\underset{m=0}{\overset{\infty}{\mathbf K}}\frac{((2m-2)z+(m-2)n)((2m+2)z+(m+2)n)}{((2m+1)(2z+n)}\end{aligned}$$
Corollaries:
1):let $z=1$ and $n=2$,then we obtain a beautiful continued fraction for square root 2 
$$\begin{aligned}{-1+\cfrac{3}{2+\cfrac{\frac{(-1)(5)}{(1)(3)}} {2+\cfrac{\frac{(1)(7)}{(3)(5)}}{2+\cfrac{\frac{(3)(9)}{(5)(7)}}{2+\cfrac{\frac{(5)(11)}{(7)(9)}}{2+\ddots}}}}}}=\sqrt{2}\end{aligned}$$
2):However the most interesting case(for me at least),occurs when we take the limit to zero 
$$\begin{aligned}\lim_{z\to0} \cfrac{z(z+1)}{2z+1+\cfrac{(0z-1)(4z+3)} {3(2z+1)+\cfrac{(2z+0)(6z+4)}{5(2z+1)+\cfrac{(4z+1)(8z+5)}{7(2z+1)+\cfrac{(6z+2)(10z+6)}{9(2z+1)+\ddots}}}}}=\frac{1}{\pi}\end{aligned}$$
yielding a new limit for $\pi$ from which one obtains the first few convergents $\begin{aligned}0,\frac{3}{8},\underline{\frac{5}{16},\frac{15}{47},\frac{7}{22}},\frac{1365}{4288},\frac{3015}{9472},\frac{1575}{4948},\ddots\end{aligned}$.
Where the underlined convergents appear in the stern-brocot tree for $\pi
$ associated to its simple continued fraction.  
Q: Is the conjectured continued fraction true (for all complex numbers $z$ with $x\gt0$)?
Update:I initially defined the continued fraction $\displaystyle\cot\left(\frac{z\pi}{4z+2}\right)$ for only natural numbers,but as a matter of fact it holds for all complex numbers $z$ with real part greater than zero.This continued fraction is one special case of the general continued fraction found here.
 A: O. Perron, Die Lehre von den Kettenbrüchen, Chapter XI, Section 82.
Satz 5 on page 488 (2nd edition, 1927):

Continued fraction
  $$
d+\underset{k=1}{\overset{\infty }{\mathbf K}}\;
\frac{a+bk+ck^2}{d+ek}
$$
  with $c \ne 0, e \ne 0, e^2+4c \ne 0$ has value
  $$
\frac{\sqrt{e^2+4c}\;{}_2F_1(\alpha,\beta;\gamma;x)\;\gamma}{{}_2F_1(\alpha+1,\beta+1;\gamma+1;x)},
$$
  where $\alpha,\beta$ are the roots of the quadratic equation
  $cZ^2-bZ+a=0$,
  $$
\gamma = \frac{b+c}{2c}\left(1-\frac{e}{\sqrt{e^2+4c}\;}\right)
+\frac{d}{\sqrt{e^2+4c}\;}
\\
x=\frac{1}{2}\left(1-\frac{e}{\sqrt{e^2+4c}\;}\right)
$$
  Choose the square-root so that
  $$
\mathrm{Re}\frac{e}{\sqrt{e^2+4c}\;}>0 .
$$

So substituting in the values in this problem, the question becomes: Is
$$
\cot\left(\frac{z\pi}{4z+2n}\right) = 
\frac{\displaystyle(z+n)\sqrt{2}\;
{}_2F_1\left(\frac{-n}{2z+n},\frac{4z+3n}{2z+n};
\frac{3}{2};\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)}
{\displaystyle(2z+n)\;
{}_2F_1\left(\frac{-(2z+2n)}{2z+n},\frac{2z+2n}{2z+n};
\frac{1}{2};
\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)} ?
$$
Writing $z/(4z+2n) = t$ and taking reciprocal, the question becomes: Is
$$
\tan(t\pi) = \frac{{}_2F_1\left(4t-2,-4t+2;
\frac{1}{2};\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)}
{\sqrt{2}(1-2t)\;{}_2F_1\left(4t-1,-4t+3;
\frac{3}{2};\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)} ?
$$
proof for this 
Use quadratic transformation

$$
{}_2F_1\left(2a,2b;a+b+\frac{1}{2};u\right)=
{}_2F_1\left(a,b;a+b+\frac{1}{2};4u(1-u)\right)
$$

to get
$$
{}_2F_1\left(4t-2,-4t+2;
\frac{1}{2};\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)
={}_2F_1\left(2t-1,-2t+1;\frac{1}{2};\frac{1}{2}\right) = \sin(\pi t)
$$
$$
{}_2F_1\left(4t-1,-4t+3;
\frac{3}{2};\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)
={}_2F_1\left(2t-\frac{1}{2},-2t+\frac{3}{2};\frac{3}{2};\frac{1}{2}\right)
=\frac{\cos(\pi t)}{\sqrt{2}(1-2t)}
$$
These are known to most CASs.
