a conjectured continued-fraction for $\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)$ that leads to a new limit for $\pi$

Solution 1:

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.