Alternative way to obtain a closed form of a continued fraction

Question

Assume the following three-term recurrence relation for $n\ge 0$ (therefore $z_{n\le 0}=0$)

\begin{equation} z_n [1 + x (2n+1)]=z_{n-1} n x +z_{n+1} (n+1) x + x (\delta_{n,0}-\delta_{n-1,0}) \end{equation} where $\delta_{m,0}$ is the kronecker-delta and $x$ is a positive parameter.

In a more compact form, \begin{equation} z_n a_n =z_{n-1} b_n +z_{n+1} c_n + Y_0(\delta_{n,0}-\delta_{n-1,0}) \end{equation}

We would like to solve this recurrence relation for $z_{0}$ taking into account all the higher terms. By this we mean the following: \begin{equation} z_0 a_0 =z_{1} c_0 + Y_0 \end{equation} \begin{equation} z_1 a_1 =z_{0} b_1 +z_{2} c_1 - Y_0 \end{equation} \begin{equation} ... \end{equation}

After some algebra, the continued fraction becomes

\begin{equation} z_0=1-\mathcal{S} \end{equation}

Being $S$ the following continued fraction

\begin{equation} \mathcal{S}=\frac{1}{a_0-\frac{c_0 b_1}{a_1-...}} \end{equation}

After some algebra manipulation and some luck, one can find that

\begin{equation} \mathcal{S}=\frac{e^{1/x}}{x}\Gamma[0,1/x] \end{equation} where $\Gamma[s,z]$ is the incomplete Gamma function which admits continued-fraction representation.

I would like to find an alternative way of solving it. The reason is that the comment I made about "after some algebra" is not easy and doesn't allow generalizations to other recurrence relations.

For that, I would like to consider the following approach given as an answer here

In the post, given a continued fraction $\mathcal{S}$, they evaluate by saying that $\mathcal{S}=\lim_{n\rightarrow \infty}\frac{P_n}{Q_n}$ and then, using the generating function, solving the ODE given by this convergence.

${\bf Progress}$: Following the idea from the link and some comments, I managed to get a recursion expression for the convergents. Define $R_{n}=(P_n, Q_n)$. Then

\begin{equation} R_{n+2}=(1+(2n+3)x)R_{n+1}-x^2 (n+1)^2 R_{n} \end{equation} for $n\ge 0$ with initial conditions $R_{0}=(0, 1)$ and $R_{1}=(a_0, 0)$.

Define $R_n=n! S_n$, then,

\begin{equation} (n+2) S_{n+2}=(1+(2n+3)x)S_{n+1}-x^2 (n+1)S_{n} \end{equation}

Define $S(z)=\sum_{\ge 0}S_n z^n$. This yields the following first-order ODE

\begin{equation} z S^{\prime}(z)[1-2xz+x^2z^2]=S(z)[(1+x)z-x^2z^2]-z(1+x)S_0+z S_1 \end{equation}

I then find that

\begin{equation} \frac{P(z)}{Q(z)}=\frac{e^{1/x}}{x}\Big(-{\rm Ei}\Big[\frac{-1}{x}\Big]+{\rm Ei}\Big[\frac{1}{x(zx-1)}\Big]\Big) \end{equation}

It is almost my result, as $-{\rm Ei}\Big[\frac{-1}{x}\Big]=\Gamma[0,1/x]$.

The only remaining part should vanish

\begin{equation} \lim_{z\rightarrow ?} {\rm Ei}\Big[\frac{1}{x(zx-1)}\Big]\rightarrow 0 \end{equation}

However, it is unclear to me how.

Answer 1

Your attempt fixups. The solution of your ODE for $S(z)$, rewritten as $$ (1-xz)^2 S'(z)-(1+x-x^2 z)S(z)=D:=S_1-(1+x)S_0, $$ with $S(0)=S_0$, is $$ S(z)=\frac1{1-xz}\exp\left(\frac1{x(1-xz)}\right)\left[S_0+D\int_0^z\frac1{1-xt}\exp\left(-\frac1{x(1-xt)}\right)\,dt\right]. $$ Assuming $x\in\mathbb{R}_{>0}$, and analysing the singularity at $z=1/x$, you should get $$ \frac{P_0+(P_1-(1+x)P_0)I}{Q_0+(Q_1-(1+x)Q_0)I}, $$ where $S_0=(P_0,Q_0)$, $S_1=(P_1,Q_1)$ (whatever these actually are) and $$ I=\int_0^{1/x}\frac1{1-xt}\exp\left(-\frac1{x(1-xt)}\right)\,dt=\frac1x\underbrace{\int_{1/x}^\infty\frac{e^{-t}}{t}\,dt}_{"=\Gamma[0,1/x]"}. $$