$LC$ ladder circuit: find the poles in a recursively defined complex sequence This question arises from considering the impedance of a ladder of inductors and capacitors, as shown in the diagram.
Suppose I have two complex numbers $Z_L(\omega) = i \omega L$ and $Z_C (\omega) = 1/i \omega C$.
I start a sequence by defining
$$Z_1(\omega) = Z_L(\omega) + Z_C(\omega) \, .$$
I then define subsequent terms in the sequence according to
$$Z_n(\omega) = Z_L(\omega) + \frac{Z_{n-1}(\omega) Z_C(\omega)}{Z_{n-1}(\omega) + Z_C(\omega)} \, .$$
I wish to analyze the position of the poles of $Z_n(\omega)$ as a function of $\omega$ for large finite $n$.
How can this be done?
Note that a related (but different) question would be: what is $\lim_{n\rightarrow \infty}Z_n(\omega)$?

Surely this could be done via induction, but one needs a guess for the recurrence relation and I have not found a way to make that guess.

 A: I assume that you want to calculate $\lim_{n \to \infty} Z_n$.
By taking limits on both sides, you find that any limit $Z$ would have to satisfy
$$Z = Z_L + \frac{Z Z_C}{Z + Z_C}.$$
Doing some algebra:
$$Z = \frac{Z_L Z + Z_L Z_C + Z Z_C}{Z + Z_C} \\
Z^2 + Z Z_C = Z (Z_L + Z_C) + Z_L Z_C \\
Z^2 - Z_L Z - Z_L Z_C = 0.$$
This is a quadratic equation whose roots are
$$Z = \frac{Z_L \pm \sqrt{Z_L^2 + 4 Z_L Z_C}}{2}.$$
So if the limit exists, it is one or the other of these two numbers. Whether the limit exists will depend on the initial condition, and when it does exist, which limit is achieved will depend on the initial condition. Usually you will converge to the possible limit which is closer to the initial condition, and there is always a domain of attraction for each limit (although it could be just the limit itself, if the limit is unstable).
To continue the calculation, it's convenient to make the variables real, so let's do that:
$$Z = \frac{iL \pm \sqrt{-L^2 + 4 L/C}}{2.}$$
Your initial condition is $i(L-1/C)$, so the difference between the roots and the initial condition is
$$Z-Z_1=\frac{iL \pm \sqrt{-L^2 + 4L/C}}{2} - i(L-1/C) = \frac{iL - 2 i L + 2 i/C \pm \sqrt{-L^2 + 4L/C}}{2} \\
= \frac{-i L + 2 i/C \pm \sqrt{-L^2 + 4L/C}}{2}.$$
This means that at least when $L/C$ is small enough, you will get the $+$ root, since
$$\frac{-iL + 2i/C + \sqrt{-L^2 + 4L/C}}{2} = i \frac{-L + 2/C + \sqrt{L^2-4L/C}}{2} \\
\approx i \frac{-L + 2/C +  L - \frac{4L/C}{2L}}{2} = 0.$$
This follows by Taylor expanding the square root. I emphasize that it requires that $L/C$ is small enough, in particular it requires that $4L/C \ll L^2$.
