@Baluncore, I think that evaluating the ladder from left to right, instead of right to left, can be better (for ##\vert b \vert<1##). A forward continued fraction like that associated with the input impedance can be evaluated in the forward direction with the Euler continuants, but there is an equivalent way of thinking which is more 'circuital' and easy to grasp, at least for me: transmission matrices.
Consider the n-th ##(n=0 \dots N)## ladder L-stage, built with a series resistance ##r_n=b^n## and a shunt conductance ##g_n=b^n##. This two-port network is described by
$$\begin{pmatrix} V_1 \\ I_1 \end{pmatrix} = T_n \begin{pmatrix} V_2 \\ I_2 \end{pmatrix}$$
with
$$T_{n} = \begin{pmatrix} 1+r_n g_n & r_n \\ g_n & 1 \end{pmatrix}$$
The T matrix of the whole N-stage ladder is
$$T = T_0 \cdot T_1 \cdot T_2 \cdots T_N$$
The key point is that the product is associative, so it can be evaluated left to right. Moreover ##T_n##
converges (for ##\vert b \vert<1##) to the identity matrix ##Id = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}## for ##n \to \infty##. This is nice since, when the arithmetic (possibly multiple) precision limit is reached, it is useless to continue multiplying: we have reached the limit value of N for that precision.
Once the ladder T matrix is computed, it is easy the find the input impedance (and other useful things, like Thevenin and Norton equivalents at the right port) as a function of the load impedance we can put to the right:
$$Z_{in} = \frac{A Z_{out}+B}{C Z_{out}+D}$$
this is one mode of getting the figure I posted in post #35.
The original
@bob012345 question has the answer
$$Z_{in}(Z_{out} = \infty) = \frac{A}{C}$$
The weird thing is, the four coefficients A, B, C and D, as (complex) functions of the (complex) variable b, converge for ##\vert b \vert<1##, and diverge for ##\vert b \vert \geq 1##. Nonetheless, ##Z_{in} = \frac{A(b)}{C(b)}## does converge for any value of b!
For ##b=1/10##, I found, with 51 stages and 50 decimal places:
$$\begin{aligned}
A &= 2.1335712819576269021587791038159481189622017101529 \\
B &= 1.2235692617434733525281261949318481910338494096970 \\
C &= 1.1223468138295548210750991909766361996746916300643 \\
D &= 1.1123458037174224855980959026243117080347869560167
\end{aligned}$$
and
$$\begin{aligned}
Z_{\infty} = Z_{in}(Z_{out} = \infty) &= 1.9009910801792873151547772738317474368474495864595 \\
Z_0 = Z_{in}(Z_{out} = 0) &= 1.0999900009999999900009999000099999999000199970002 \\
Z_1 = Z_{in}(Z_{out} = 1) &= 1.5022829168273864624216950620116492911165292953097
\end{aligned}$$
An interesting fact is that ##Z_{in}(Z_{out})## can be written as:
$$Z_{in} = \frac{Z_{\infty}(Z_1-Z_0) Z_{out}+Z_0 (Z_{\infty}-Z_1)}{(Z_1-Z_0)Z_{out}+(Z_{\infty}-Z_1)}$$