For what $x\in\mathbb{R}\setminus\{0\}$ does generalized continued fraction $x+\frac{x}{x+\frac{x}{x+\frac{x}{x+...}}}$ converge? Being an undergraduate student knowing only information from Wikipedia on the subject, I tried to find an non-recursive formula for the sequence of convergents $x,\frac{x(x+1)}{x},\frac{x^2(x+2)}{x(x+1)},...$ but could not do so.
What approaches can one take to demonstrate convergence or divergence of this generalized continued fraction? When it converges, how can one compute the limit?
Condition for Convergence:
Let $k = x+\frac{x}{x+\frac{x}{x+\ldots}}$.
It follows that $k = x + \frac{x}{k}$.
Solve the equation for $k$.
$$k^2-xk-x=0$$ $$k=\frac{x\pm\sqrt{x^2+4x}}{2}$$
$k$ converges only if $x^2+4x\geq0$, i.e. $\underline{\underline{x\leq-4 \text{ or } x>0}}$.
Limit
When $x>0$, it is apparent that $k>0$, so $\underline{\underline{k=\dfrac{x+\sqrt{x^2+4x}}{2}}}$ because $x<\sqrt{x^2+4x}$.
When $x\leq-4$, suppose $k>0$. Then by observing $k=x+\frac{x}{k}$, contradiction arises because $\text{LHS}>0\text{ and RHS}<0$.
Therefore, $k<0$. And $\underline{\underline{k=\dfrac{x-\sqrt{x^2+4x}}{2}}}$.