Convergence of generalised continued fractions (with positive partial numerators) Suppose that we have a sequence of positive numbers $(x_n)_{n \in \mathbb N}: x_n>0$ which are not necessarily integers.
Q1 Can you give some examples of necessary/sufficient conditions for the convergents of the generalised continued fraction expansion associated to $(x_i)$, given by (to use Gauss's notation:)
$$
  x_1 + \underset{n=2}{\overset\infty {K}}\; \frac 1 {x_n} \quad=\quad  x_1 + \frac 1 {x_2 + \frac 1 {x_3 + \ddots}},
$$
to converge?
Q2 On the other hand, what can possibly go wrong—so as to guarantee divergence?
References are certainly welcome. Thanks for reading!
 A: A reference is Analytic Theory of Continued Fractions by H. S. Wall, American Mathematical Society, Chelsea Publications, reprint 2000.  This is quite old (first published 1948) so the subject may have advanced since.  See Chapter II and III for the following theorems for complex arguments $b_n$ and the continued fraction,
$$\begin{align}   \dfrac{1}{b_1+\dfrac{1}{b_2+\dfrac{1}{b_3+\cdots}}} \tag 1 \end{align}$$


*

*If the series $\sum b_{2n}$ and $\sum b_{2n+1} $ converge and at least one is absolutely convergent then the continued fraction $(1)$ is divergent (von Koch, p.33).

*If $\lvert b_n \rvert > 2 $ then $(1)$ converges (Pringsheim, p.50 slightly adapted).

*If the $b_n$ satisfy
$$ \begin{align} & \Re{(b_1)} > 0, \text{ and } \Re{(b_k)} \geqslant 0, k=2,3,\cdots~,\\
& \sum{\lvert b_{2n+1} \rvert } \text{ converges}~, \\
& \sum \lvert b_{2n+1}s_n^2 \rvert  \text{ converges, where } s_n = b_2 +b_4+\cdots+b_{2n}~, \text{ and}, \\
&\lim \lvert s_n \rvert = \infty \end{align}$$
then the continued fraction $(1)$ converges (Scott and Wall, p.34).
I am not aware that there are any universal convergence / divergence criteria.   The conditions above are sufficient but not necessary.  
A: At https://en.wikipedia.org/wiki/Convergence_problem there is a discussion of convergence for generalized continued fractions.
Consider the sequence of partial denominators: $q_1=1,\,q_2=x_2,\,q_n=x_n\cdot q_{n-1}+q_{n-2},\forall n\geq 3$. Then the continued fraction in question converges if, and only if, $\displaystyle\lim_nq_{n+1}q_n\to \infty$.
This can be proved using the difference formula:
$[x_1,x_2,\dots,x_n]-[x_1,x_2,\dots,x_n,x_{n+1}]=\dfrac{(-1)^{n+1}}{q_nq_{n+1}}$.
And observing that, since $x_i$ is positive for every $i$, the continued fraction $[x_1,x_2,\dots,x_n,x_{n+1},\dots,x_{n+a}]$ is in between $[x_1,x_2,\dots,x_n]$ and $[x_1,x_2,\dots,x_n,x_{n+1}]$ for every $a\geq 2$.
As observed by WA Don in the comments this convergence criteria can be further refined. Actually, the continued fraction in question converges if, and only if, $\displaystyle\sum_k x_k=\infty$.
According to Jones and Thron, Continued Fractions (1980), this can be proved using the inequalities $q_n\leq (1+x_2)\cdots (1+x_n)\leq e^{\sum_{k=2}^nx_k}$ for every $n$ and $q_{2k}\geq x_2+x_4+\cdots x_{2k}$ and $q_{2k+1}\geq 1+x_2(x_3+x_5+\cdots+x_{2k+1})$ for every $k$.
In the particular case where $x_k$ is uniformly bounded away from zero, i.e. $c:=\inf_k x_k>0$, then the convergence of the continued fraction is exponential. This follows from the inequality $q_{n+1}q_n\geq c(1+c^2)^{n-1}$ for every $n$.
