By iterating the basic relation $$ \forall z \in \mathbb{C} \setminus \{ -1 \}: \quad z = 1 + \frac{z^{2} - 1}{z + 1}, $$ one obtains the following finite continued-fraction identities: \begin{alignat}{2} \forall z \in \mathbb{C} \setminus \{ -1,-2 \}: & \quad & z & = 1 + \frac{z^{2} - 1}{1 + \dfrac{(z + 1)^{2} - 1}{z + 2}}, \\ \forall z \in \mathbb{C} \setminus \{ -1,-2,-3 \}: & \quad & z & = 1 + \frac{z^{2} - 1} {1 + \dfrac{(z + 1)^{2} - 1}{1 + \dfrac{(z + 2)^{2} - 1}{z + 3}}}, \\ \vdots \end{alignat} I was therefore led to speculate that for every $ z \in \mathbb{C} \setminus \mathbb{Z}_{< 0} $, the following infinite continued-fraction identity holds (written using Ramanujan Notation): \begin{align} z & = 1 + \frac{z^{2} - 1}{1 +} \frac{(z + 1)^{2} - 1}{1 +} \frac{(z + 2)^{2} - 1}{1 +} \cdots \\ & = \lim_{n \to \infty} \left[ 1 + \frac{z^{2} - 1}{1 +} \cdots \frac{(z + n)^{2} - 1}{1} \right]. \end{align} Unfortunately, I was unable to supply a rigorous proof of this identity.
Perhaps someone could kindly offer some suggestions. Thank you very much!
With $a(n)=(z+n-1)^2-1$ and $b(n)=z+n,$ the first two identities are $$z=1+\frac{a(1)}{b(1)}, \\ z=1+\frac{a(1)}{1+\frac{a(2)}{b(2)}}.$$ For the continued fraction convergents, the final denominator of $b(n)$ is replaced by $1,$ and so to check convergence we look at the differences $f(n)=C(n)-z$ where $C(n)$ is the $n$th convergent. I found $$f(1)=z^2-z,\\ f(2)=\frac{z-z^2}{z+1}, \\ f(3)=\frac{z^2-z}{2}, \\ f(4)=\frac{z-z^2}{z+2}, \\ f(5)=\frac{z^2-z}{3}, \\ f(6)=\frac{z-z^2}{z+3}.$$ The pattern is that, for odd $n=2k-1,$ $f(n)=\frac{z^2-z}{k},$ while for even $n=2k$ we have $f(n)=\frac{z-z^2}{z+k}.$
I don't know how to base an induction to establish these hold for all $k,$ maybe one would need some general results on computing convergents for a continued fraction of this type, where the "numerators" are not all $1.$ However if these formulas are right, one can see the continued fraction converges to $z$ as long as $z$ is not a negatived integer. [Added later, see below under "a better way...", I found a way to inductively establish the forms for the $C(n)$ so this part is finished.] For odd $n$ it is clear, while for even $n$ the denominator norm $|z+k|$ can be bounded below by $|z+k| \ge ||z|-k|,$ which for fixed $z$ goes to $+\infty$ as $k \to \infty,$ so one can conclude that indeed $f(n) \to 0$ as $n \to \infty$ and the continued fraction converges to $z.$
A better way via explicit formulas for the convergents.
If as above $C(n)$ is the $n$th convergent, then one can establish by induction the following two formulas, based on the parity of $n.$ $$C(2k-1)=\frac{z^2+(k-1)z}{k},\\ C(2k)=\frac{kz+z}{k+z}.\tag{1}$$ Since each of these goes to $z$ as $k \to \infty$ the convergence of the continued fraction follows without need of subtracting the limit $z$ from them.
The method for doing the induction step is to define $C(k,z)$ as the $k$ convergent if the starting number is $z,$ and note that for any positive $r$ we have $$C(r,z)=1+\frac{z^2-1}{C(r-1,z+1)},$$ by the way the convergents are defined. This "peels off" the top $1+(z^2-1)$ part, so that there are $r-1$ layers remaining but $z$ has increased by $1.$ It is then routine to get from the case $r=2k-1$ to $r=2k,$ and then from $r=2k$ to $r=2k+1.$ In each case we apply the previous formula from $(1)$ as appropriate, with the $z$ values adjusted, and just check the algebra.