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 $\displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)$ is true

$$\begin{split}\displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)&=\frac{\displaystyle\Gamma\left(\frac{z+n}{4z+2n}\right)\Gamma\left(\frac{3z+n}{4z+2n}\right)}{\displaystyle\Gamma\left(\frac{z}{4z+2n}\right)\Gamma\left(\frac{3z+2n}{4z+2n}\right)}\\&=\cfrac{2z}{2z+n+\cfrac{(n)(4z+n)} {3(2z+n)+\cfrac{(2z+2n)(6z+2n)}{5(2z+n)+\cfrac{(4z+3n)(8z+3n)}{7(2z+n)+\ddots}}}}\end{split}$$


By taking the limit(which follows after abel's theorem) $$ \begin{aligned}\lim_{z\to0}\frac{\displaystyle\tan\left(\frac{z\pi}{4z+2}\right)}{2z}=\frac{\pi}{4}\end{aligned}, $$ we recover the well known continued fraction for $\pi$


If we let $z=1$ and $n=2$,then we have the square root of $2$ $$\begin{aligned}{1+\cfrac{1}{2+\cfrac{1} {2+\cfrac{1}{2+\cfrac{1}{2+\ddots}}}}}=\sqrt{2}\end{aligned}$$

Q: How do we prove rigorously that the conjectured continued fraction is true and converges for all complex numbers $z$ with $x\gt0$?

Update:I initially defined the continued fraction $\displaystyle\tan\left(\frac{z\pi}{4z+2}\right)$ for only natural numbers,but as a matter of fact it holds for all complex numbers $z$ with real part greater than zero.Moreover,this continued fraction is a special case of the general continued fraction found in this post.

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    $\begingroup$ That is impressive. How did you come up with it? $\endgroup$ Commented Oct 13, 2015 at 16:20
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    $\begingroup$ Really, how did you get this? If you provide some of the methods you used, there is a better chance someone would be able to answer your question. $\endgroup$
    – Yuriy S
    Commented Mar 13, 2016 at 19:01
  • $\begingroup$ @Nicco, thank you for the link. Sorry, but I don't have anything to contribute so far $\endgroup$
    – Yuriy S
    Commented Apr 7, 2016 at 17:04
  • $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ Commented Mar 12, 2018 at 17:02

2 Answers 2


The proposed continued fraction \begin{equation} \displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)=\cfrac{2z}{2z+n+\cfrac{(n)(4z+n)} {3(2z+n)+\cfrac{(2z+2n)(6z+2n)}{5(2z+n)+\cfrac{(4z+3n)(8z+3n)}{7(2z+n)+\ddots}}}} \end{equation} can be written as \begin{equation} \displaystyle\tan\left(\frac{z\pi}{4z+2n}\right)=\cfrac{2z/\left( 2z+n \right)}{1+\cfrac{(n)/\left( 2z+n \right)\cdot(4z+n)/\left( 2z+n \right)} {3+\cfrac{(2z+2n)/\left( 2z+n \right)\cdot(6z+2n)/\left( 2z+n \right)}{5+\cfrac{(4z+3n)/\left( 2z+n \right)\cdot(8z+3n)/\left( 2z+n \right)}{7+\ddots}}}} \end{equation} Denoting $u=\cfrac{z}{4z+2n}$, the factors of the numerators are \begin{equation} \frac{n}{2z+n}=1-4u\,;\quad\frac{4z+n}{2z+n}=1+4u\,;\quad\frac{2z+2n}{2z+n}=2-4u\,;\quad\frac{6z+2n}{2z+n}=2+4u\,;\cdots \end{equation} Then, the fraction can be simplified as \begin{equation} \displaystyle\tan\left(\pi u\right)=\cfrac{4u}{1+\cfrac{\cfrac{1-16u^2}{1\cdot3}} {1+\cfrac{\cfrac{4-16u^2}{3\cdot5}}{1+\cfrac{\cfrac{9-16u^2}{5\cdot7}}{1+\ddots}}}} \end{equation} It is thus a special case of the continued fraction found in this answer: \begin{equation} \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}}}} \end{equation} here $z=1$ and $\alpha=4u$. The brilliant proof is based on a continued fraction due to Nörlund.


The ratio $$\tan\dfrac{\pi z}{4z+2n} = \dfrac{\Gamma\left(\dfrac{z+n}{4z+2n}\right)\Gamma\left(\dfrac{3z+n}{4z+2n}\right)}{\Gamma\left(\dfrac{z}{4z+2n}\right)\Gamma\left(\dfrac{3z+2n}{4z+2n}\right)}\hspace{100mu}\tag1$$ can be obtained, applying "real" identity

$$\sin\pi x = \dfrac\pi{\Gamma(x)\Gamma(1-x)}\hspace{100mu}\tag2$$

to the expression $$\tan\dfrac\pi2\dfrac z{2z+n} = \dfrac{\sin\pi\dfrac z{4z+2n}}{\sin\pi\dfrac{z+n}{4z+2n}},$$ so it looks nice and quite correct.

Continued fraction can be obtained, using known continued fraction of the tangent function in the form of $$\tan \dfrac{\pi x}4 = \cfrac x{1+\operatorname{ \Large K}\hspace{-27mu}\phantom{\Big|}_{k=1}^{\large ^{\,\infty}}\cfrac{(2k-1)^2-x^2}2}\hspace{100mu}\tag3$$ with $$x=\dfrac{2z}{2z+n}.$$


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