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Given complex numbers $a=x+iy$, $b=m+in$ and a gamma function $\Gamma(z)$ with $x\gt0$ and $m\gt0$, it is conjectured that the following continued fraction holds

$$\frac{\displaystyle4\Gamma\left(\frac{2a+3}{4}\right)\Gamma\left(\frac{2b+3}{4}\right)}{\displaystyle\Gamma\left(\frac{2a+1}{4}\right)\Gamma\left(\frac{2b+1}{4}\right)}=\cfrac{(2a+1)(2b+1)}{a+b+2+\cfrac{(a-b+1)(b-a+1)} {a+b+4+\cfrac{(2a+3)(2b+3)}{a+b+6+\cfrac{(a-b+3)(b-a+3)}{a+b+8+\ddots}}}}\tag{1a}$$

Corollary

$$\frac{4}{\pi}=\cfrac{(2)^2}{3+\cfrac{(1)^2}{5+\cfrac{(4)^2}{7+\cfrac{(3)^2}{9+\ddots}}}}\tag{1b}$$

Q: How do we prove the continued fraction $(1a)$ rigorously?

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  • $\begingroup$ Equation (1b) is wrong. The left side should be $1/\pi.$ For example, $1/(3+1/5)=5/16=0.3125$ while $1/\pi\approx 0.3183.$ $\endgroup$ – Somos May 20 at 19:05
  • $\begingroup$ @Somos : I doubt it is wrong. It likely has a very slow rate of convergence. I'll post some of the reasoning that led me to that conjecture shortly $\endgroup$ – Nicco May 21 at 19:22
  • $\begingroup$ I see the problem. A bit of calculation shows that the left and right side of your first equation multiplied is $4$. Thus, when $a=b=1/2$ the left side gamma quotient is $\pi$ and the right side continued fraction is $4/\pi.$ $\endgroup$ – Somos May 21 at 20:25
  • $\begingroup$ @Somos : Yes you're right, a bit of calculation does indicate that the second continued fraction converges to $\frac{1}{\pi}$ $\endgroup$ – Nicco May 21 at 22:42
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    $\begingroup$ The third continued fraction in the Wikipedia article for Pi is similar to yours. See the image here. $\endgroup$ – Somos May 27 at 14:07
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This can be deduced from Gauss' continued fraction $$\frac{_2F_1(a+1,b;c+1;z)}{_2F_1(a,b;c;z)}=\cfrac{c}{c+\cfrac{(a-c)bz}{c+1+\cfrac{(b-c-1)(a+1)z}{c+2+\cfrac{(a-c-1)(b+1)z}{c+3+\cfrac{(b-c-2)(a+2)z}{c+4+\ddots}}}}}$$ which evaluates the continued fraction in $(1a)$ as $$\frac{(2a+1)(2b+1)}{a+b+2}\frac{_2 F_1\left(b+\frac32,\frac{b-a+1}{2};\frac{a+b}{2}+2;-1\right)}{_2 F_1\left(b+\frac12,\frac{b-a+1}{2};\frac{a+b}{2}+1;-1\right)},$$ and Kummer's formula $_2 F_1(a,b;a-b+1;-1)=\dfrac{\Gamma(a/2+1)\Gamma(a-b+1)}{\Gamma(a+1)\Gamma(a/2-b+1)}$.

After all the necessary substitutions and cancellations, this yields exactly the expected result. The proofs are sketched in the linked articles, but you may want to follow further references for a deeper treatment.

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