Sum of an infinite series with continued fractions

Question

I'm trying to determine if it is possible to sum an infinite series of the form:

\begin{equation} \frac{1}{x}+\frac{a}{x^2(x-\frac{a}{x})}+\frac{a^2}{x^2(x-\frac{a}{x})^2(x-\frac{a}{x-\frac{a}{x}})}+\frac{a^3}{x^2(x-\frac{a}{x})^2(x-\frac{a}{x-\frac{a}{x}})^2(x-\frac{a}{x-\frac{a}{x-\frac{a}{x}}})}+\ldots \end{equation}

where you can see that each successive term can be obtained from the previous one, by multiplying by $a$, squaring one of the terms in the denominator and adding a continued fraction with an extra contribution.

I know how to sum the infinite continued fraction that one gets for the "last" term, but I do not know how could I sum the whole expression, or even if it is possible. I have checked that numerically it converges, at least for a certain range.

Answer 1

A simple question unanswered for such a long time. I'm assuming $a,x\in\mathbb{R}$ and $a\neq 0$.

Let $x_0=x$ and $x_{n+1}=x-a/x_n$; then the question asks for $\sum_{n=0}^\infty y_n$, where $y_0=1/x$ and $y_{n+1}=(ay_n)/(x_n x_{n+1})$. But we have $\color{blue}{\sum_{n=0}^m y_n=1/x_m}$ (easy to check using induction on $m$). Hence the given sum converges if and only if the sequence $(x_n)$ converges (and is well-defined).

The latter happens if and only if $\color{blue}{x^2\geqslant 4a}$; this (together with the limit itself) can be deduced from the closed form $x_n=z_{n+2}/z_{n+1}$, where $z_n=(\lambda_1^n-\lambda_2^n)/(\lambda_1-\lambda_2)$ and $\lambda_1,\lambda_2$ are the (possibly complex) roots of $\lambda^2-x\lambda+a=0$; in the case $\lambda_1=\lambda_2=\lambda$ we take $z_n=n\lambda^{n-1}$.

Answer 2

Just trying to understand how metamorhy obtained his answer... we define:

$y_m \equiv \frac{1}{x_m}-\frac{1}{x_{m-1}} \ (0)$

and we show that this definition respects the (recursive) definition of metamorphy:

$y_{m+1}=\frac{y_ma}{x_mx_{m+1}}\ (1)$.

To show Eq. (1) observe that from the definition $(0)$ we obtain:

$\frac{y_{m+1}}{y_m}=\frac{x_m-x_{m+1}}{x_{m-1}-x_{m}} \frac{x_{m-1}}{x_{m+1}} \ (2)$

Substracting the equations $x_{m+1}=x-a/x_m$ and $x_{m}=x-a/x_{m-1}$ we have that $\frac{x_m-x_{m+1}}{x_{m-1}-x_{m}}=\frac{a}{x_mx_{m-1}}$. Substituting this relation in (2) we obtain (1), which is the definition/relation that metamorphy used.

Problem is, here I started from (0) and obtained (1), whereas metamorphy did the opposite. So either he guessed the result ( Eq. (0)), or he followed a different set of manipulations to start from (1), the Eq. that sounds more natural to start with the problem, and obtain (0), which is what makes the sum a telescopic one...