Continued fractions

Continued fraction expansion for $\coth\left(z\right)$

We will show the proof for this continued fraction posted by @infseriesbot. \[ \frac{e+1}{e-1} = \coth\left(\frac{1}{2}\right) = 2+\cfrac{1}{6+\cfrac{1}{10+\cfrac{1}{14+\cdots}}}\] Update We add the proof of this result, which is bascially the same function evaluated at other point: \[ 0 = \coth\left(-\frac{i\pi}{2}\right) = 2-\cfrac{\pi^2}{6-\cfrac{\pi^2}{10-\cfrac{\pi^2}{14-\cdots}}}\]
Recall the continued fraction representation for ratios of confluent hypergeometric functions of the type $_{0}F_{1}(c,z)$: Let $(a_{n})$ be a sequence of complex numbers defined by \[a_{n} = \frac{1}{(c+n-1)(c+n)} \] where $c$ is a complex constant such that $c\notin \mathbb{Z^{-}}\cup \{0\} $. Then \[1+\mathcal{K}_{n=1}^{\infty} \left(\frac{a_{n}z}{1}\right) = \frac{_{0}F_{1} (c,z)}{_{0}F_{1} (c+1,z)}\]
Now, for this particular case, we will use the hypergemetric representation of $\sin(z)$ and $\cos(z)$ \[\sin(z)= z \, {_{0}F_{1}}\left(\frac{3}{2},-\frac{z^2}{4}\right)\] \[\sin(z)= {_{0}F_{1}}\left(\frac{1}{2},-\frac{z^2}{4}\right)\] Then \begin{align*} \tan(x) = &\frac{\sin(x)}{\cos(x)} = \frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...}{1+\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...}\\ =& \frac{x(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+...)}{1+\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...}\\ =& \frac{x\cdot _{0}F_{1}\left(\frac{3}{2},-\frac{x^2}{4}\right)}{ _{0}F_{1}\left(\frac{1}{2},-\frac{x^2}{4}\right)}\\ =& \frac{x}{1-\cfrac{x^2}{3-\cfrac{x^2}{5-\cfrac{x^2}{7-\cdots}}}} \end{align*} If we evaluate at $x=\frac{iz}{2}$ \begin{align*} \tan\left(\frac{iz}{2}\right) = &\frac{iz^2}{2-\cfrac{-z^2}{6-\cfrac{-z^2}{10-\cfrac{-z^2}{14-\cdots}}}}\\ =&\frac{iz}{2+\cfrac{z^2}{6+\cfrac{z^2}{10+\cfrac{z^2}{14-\cdots}}}}\\ \Longrightarrow \coth\left(\frac{z}{2}\right)z = &\cot\left(\frac{iz}{2}\right)iz = 2+\cfrac{z^2}{6+\cfrac{z^2}{10+\cfrac{z^2}{14+\cdots}}} \end{align*} If we evaluate at $z=1$ \[ \boxed{ \frac{e+1}{e-1} = \coth\left(\frac{1}{2}\right) = 2+\cfrac{1}{6+\cfrac{1}{10+\cfrac{1}{14+\cdots}}}}\] Since the result is valid for al the complex plane, if we evaluate at $z=-i\pi$ then $\displaystyle \coth\left(-\frac{i\pi}{2}\right) =0$ and \[\boxed{ 0 = \coth\left(-\frac{i\pi}{2}\right) = 2-\cfrac{\pi^2}{6-\cfrac{\pi^2}{10-\cfrac{\pi^2}{14-\cdots}}}}\]