双曲線関数の逆数和の連分数展開

$$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}\frac{z^n}{a{r}^n-b{q}^n}& =& \underset{n=0}{\overset{\mathrm{\infty }}{\begin{array}{c}\mathrm{K}\end{array}}}\frac{v_n}{a{r}^n-b{q}^n}\end{array}$$
$$\begin{array}{r}\{\begin{array}{lll}{v}_0& =& 1\\ v_{2n-1}& =& -(rq{)}^{n-1}{(a{r}^{n-1}-b{q}^{n-1})}^{2}z\\ v_{2n}& =& -ab(rq{)}^{n-1}{(r^n-q^n)}^{2}z\end{array}\end{array}$$

$$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}\frac{z^n}{{\cosh }_{\pm }(an+b)}& =& 2\sum _{n=0}^{\mathrm{\infty }}\frac{z^n}{e^{an+b}\pm e^{-an-b}}\\ & =& \underset{n=0}{\overset{\mathrm{\infty }}{\begin{array}{c}\mathrm{K}\end{array}}}\frac{v_n}{{\cosh }_{\pm }(an+b)}\end{array}$$
$$\begin{array}{r}\{\begin{array}{lll}{v}_0& =& z\\ v_{2n-1}& =& -{\cosh }_{\pm }^2(a(n-1)+b)z\\ v_{2n}& =& \pm {\sinh }^2(an)z\end{array}\end{array}$$