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.

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.
