conjectured general continued fraction for the quotient of gamma functions

Continued fraction (2) can be simplified as $$ \tan\left(\alpha\tan^{-1}z\right)=\cfrac{\alpha z}{1+\cfrac{\frac{(1^2-\alpha^2)z^2}{1\cdot 3}} {1+\cfrac{\frac{(2^2-\alpha^2)z^2}{3\cdot 5}}{1+\cfrac{\frac{(3^2-\alpha^2)z^2}{5\cdot 7}}{1+\ddots}}}}\tag{2a} $$ This is a special case of the following continued fraction due to N$\ddot{\text{o}}$rlund (B.Berndt, Ramanujan's notebooks, vol.2)

To obtain (2a) from (21.6) set $\beta=-\alpha,\gamma=\frac{1}{2},\frac{x(1-x)}{(1/2-x)^2}=z^2$. Now one can apply the following formulas $$ \, _2F_1\left(a,-a;\frac{1}{2};x\right)=\cos \left(2 a \sin ^{-1}\left(\sqrt{x}\right)\right) $$ $$ -2a^2\, _2F_1\left(1+a,1-a;\frac{3}{2};x\right)=\frac{d}{dx}\left[\, _2F_1\left(a,-a;\frac{1}{2};x\right)\right] $$ $$ 2 \sin ^{-1}\left(\sqrt{x}\right)=\tan^{-1}\frac{\sqrt{x(1-x)}}{1/2-x} $$ to complete the proof.

SECTION 1.

I use the identity $$ \Gamma(z)\Gamma(1-z)=\pi \csc(\pi z)\textrm{, }z\in\textbf{C}-\textbf{Z} $$ to get $$ \Gamma\left(\frac{a+3b}{4(a+b)}\right)\Gamma\left(\frac{3a+b}{4(a+b)}\right)=\pi\csc\left(\pi\frac{a+3b}{4(a+b)}\right), \tag 1 $$ since $$ \frac{a+3b}{4(a+b)}+\frac{3a+b}{4(a+b)}=1. $$ Using the same argument and the relation $$ \Gamma(1+z)=z\Gamma(z)\textrm{, }Re(z)>0 $$ I can write $$ \Gamma(z)\Gamma(2-z)=(1-z)\Gamma(z)\Gamma(1-z), $$ and hence $$ \Gamma\left(\frac{3a+5b}{4(a+b)}\right)\Gamma\left(\frac{5a+3b}{4(a+b)}\right)=\left(1-\frac{3a+5b}{4(a+b)}\right)\pi\csc\left(\pi\frac{3a+5b}{4(a+b)}\right)= $$ $$ =\frac{a-b}{4(a+b)}\pi\csc\left(\pi\frac{3a+5b}{4(a+b)}\right)\tag 2 $$ Hence $$ \frac{\Gamma\left(\frac{a+3b}{4(a+b)}\right)\Gamma\left(\frac{3a+b}{4(a+b)}\right)}{\Gamma\left(\frac{3a+5b}{4(a+b)}\right)\Gamma\left(\frac{5a+3b}{4(a+b)}\right)}=\frac{4(a+b)}{a-b}\frac{\csc\left(\pi\frac{a+3b}{4(a+b)}\right)}{\csc\left(\pi\frac{3a+5b}{4(a+b)}\right)}=\frac{4(a+b)}{a-b}\frac{\sin\left(\pi\frac{3a+5b}{4(a+b)}\right)}{\sin\left(\pi\frac{a+3b}{4(a+b)}\right)}= $$ $$ =\frac{4(a+b)}{a-b}\frac{\sin\left(\pi\frac{3a+5b}{4(a+b)}\right)}{\sin\left(\pi\frac{a+3b}{4(a+b)}\right)}=4\frac{a+b}{a-b}\frac{\cos\left(\pi\frac{a+3b}{4(a+b)}\right)}{\sin\left(\pi\frac{a+3b}{4(a+b)}\right)}= $$ $$ =4\frac{a+b}{a-b}\cot\left(\pi\frac{a+3b}{4(a+b)}\right)=4\frac{a+b}{a-b}\cot\left(\pi\frac{a-b}{4(a+b)}\right)=4\frac{a+b}{a-b}\tan\left(\pi\frac{a-b}{4(a+b)}\right).\tag 3 $$ From the reason that $\frac{3a+5b}{4(a+b)}-\frac{a+3b}{4(a+b)}=\frac{1}{2}$ and $$ \cot\left(\pi\frac{a+3b}{4(a+b)}\right)=\tan\left(\frac{2(a+b)\pi}{4(a+b)}-\pi\frac{a+3b}{4(a+b)}\right). $$ The conditions are $$ \frac{a+3b}{4(a+b)}\textrm{, }\frac{3a+5b}{4(a+b)}\not\in\textbf{Z}\textrm{ and }Re\left(\frac{a-b}{a+b}\right)>0\tag 4 $$

NOTE 1.

The $n-$th convergent of the numerator of Gamma-function-involved cf can written as $$ \left((n+1)a+(n-1)b\right)\left((n-1)a+(n+1)b\right)=n^2(a+b)^2-(a-b)^2 $$

SECTION 2.

In this section we prove the generalization (second cf stated in the answer). Consequently from this cf and Section 1 we get the validity of the first stated cf (this with the Gamma-function).

The following expansion is at [1] p.566 $$ -1+\left(\frac{1+Az}{1+Bz}\right)^\alpha= $$ $$ \frac{2\alpha (A-B)z}{2+(A+B-\alpha (A-B))z-}\frac{(A-B)^2(1^2-\alpha^2)z^2}{3(2+(A+B)z)-}\frac{(A-B)^2(2^2-\alpha^2)z^2}{5(2+(A+B)z)-}\frac{(A-B)^2(3^2-\alpha^2)z^2}{7(2+(A+B)z)-\ldots} $$ for $$ \frac{2+(A+B)z}{(A-B)z}\in\textbf{C}-[-1,1].\tag 4 $$

Setting $$ A=\frac{-2+(a+b)(d-i\sqrt{c})}{2z}\textrm{, }B=\frac{-2+(a+b)(d+i\sqrt{c})}{2z}\textrm{ and } \alpha=\frac{b-a}{b+a} $$ we get $$ \frac{2\alpha(A-B)z}{\left(\frac{1+Az}{1+Bz}\right)^{\alpha}-1}-\left(2+(A+B-\alpha(A-B))z\right)=\ldots= $$ $$ =\frac{b-a}{c^{-1/2}\tan\left(\frac{b-a}{b+a}\arctan\left(\frac{\sqrt{c}}{d}\right)\right)}-d(a+b) $$ Since also $$ -(A-B)^2(n^2-\alpha^2)z^2=c\left((a+b)^2n^2-(a-b)^2\right) $$ and $$ 2+(A+B)z=d(a+b), $$ we get the proof for the generalization cf. The condition (4) then is $$ \frac{id}{\sqrt{c}}\in\textbf{C}-[-1,1]. $$ Observe that, the "generalization cf" for $c=d=1$ is always convergent and the convergent function is (easily): $$ 4\frac{a+b}{a-b}\tan\left(\pi\frac{a-b}{4(a+b)}\right) $$ Hence from Section 1 (relation (3)) we get the result with conditions (4).

[1]: Lisa Lorentzen, Haakon Waadeland."Continued Fractions with Applications" . North-Holland.,Amsterdam, London, New York, Tokyo. (1992).